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Physics 5th Edition Walker Solutions Manual
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Chapter 2: One-Dimensional Kinematics
Answers to Even-Numbered Conceptual Questions
2.
An odometer measures the distance traveled by a car. You can tell this by the fact that an odometer has a nonzero
reading after a round trip.
4.
No. Their velocities are different because they travel in different directions.
6.
Since the car circles the track its direction of motion must be changing. Therefore, its velocity changes as well. Its
speed, however, can be constant.
8.
(a) The time required to stop is doubled. (b) The distance required to stop increases by a factor of four.
10.
Yes, if it moves with constant velocity.
12.
(a) No. If air resistance can be ignored, the acceleration of the ball is the same at each point on its flight. (b) Same
answer as part (a).
14.
(a) No. Displacement is the change in position, and therefore it is independent of the location chosen for the origin. (b)
Yes. In order to know whether an objectโs displacement is positive or negative, we need to know which direction has
been chosen to be positive.
Solutions to Problems and Conceptual Exercises
1.
Picture the Problem: You walk in both the positive and negative
directions along a straight line.
Strategy: The distance is the total length of travel, and the
displacement is the net change in position. We place the origin at
the location labeled โYour house.โ
Solution: 1. (a) Add the lengths:
๏จ 0.75 ๏ซ 0.60 mi ๏ฉ ๏ซ ๏จ 0.60 mi ๏ฉ ๏ฝ 1.95 mi
2. (b) Subtract xi from xf to find the displacement.
๏ x ๏ฝ xf ๏ญ xi ๏ฝ 0.75 ๏ญ 0.00 mi ๏ฝ 0.75 mi
Insight: The distance traveled is always positive, but the displacement can be negative.
2.
Picture the Problem: You walk in both the positive and negative
directions along a straight line.
Strategy: The distance is the total length of travel, and the
displacement is the net change in position. We place the origin at
the location labeled โYour house.โ
Solution: 1. (a) Add the lengths:
๏จ 0.60 ๏ซ 0.35 mi ๏ฉ ๏ซ ๏จ 0.75 ๏ซ 0.60 ๏ซ 0.35 mi ๏ฉ ๏ฝ 2.65 mi
2. (b) Subtract xi from xf to find the displacement.
๏ x ๏ฝ xf ๏ญ xi ๏ฝ 0.00 ๏ญ 0.75 mi ๏ฝ ๏ญ 0.75 mi
Insight: The distance traveled is always positive, but the displacement can be negative.
Copyright ยฉ 2017 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No
portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
2โ1
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James S. Walker, Physics, 5th Edition
Chapter 2: One-Dimensional Kinematics
3.
Picture the Problem: Player A walks in the positive direction and player B walks
in the negative direction.
Strategy: In each case the distance is the total length of travel, and the
displacement is the net change in position.
Solution: 1. (a) Note the distance traveled by player A:
5m
2. The displacement of player A is positive:
๏ x ๏ฝ xf ๏ญ xi ๏ฝ 5 m ๏ญ 0 m ๏ฝ 5 m
3. (b) Note the distance traveled by player B:
2m
4. The displacement of player B is negative. Let
the origin be at the initial position of player A.
๏ x ๏ฝ xf ๏ญ xi ๏ฝ 5 m ๏ญ 7 m ๏ฝ ๏ญ2 m
Insight: The distance traveled is always positive, but the displacement can be negative.
4.
Picture the Problem: The ball is putted in the positive direction
and then the negative direction.
Strategy: The distance is the total length of travel, and the
displacement is the net change in position.
Solution: 1. (a) Add the lengths:
๏จ10 ๏ซ 2.5 m๏ฉ ๏ซ 2.5 m ๏ฝ 15 m
2. (b) Subtract xi from xf to find the displacement.
๏ x ๏ฝ xf ๏ญ xi ๏ฝ 10 ๏ญ 0 m ๏ฝ 10 m
Insight: The distance traveled is always positive, but the displacement can be negative.
5.
Picture the Problem: The runner moves along the oval track.
Strategy: The distance is the total length of travel, and the
displacement is the net change in position.
Solution: 1. (a) Add the lengths:
๏จ15 m๏ฉ ๏ซ ๏จ100 m๏ฉ ๏ซ ๏จ15 m๏ฉ ๏ฝ 130 m
2. Subtract xi from xf to find the displacement.
๏ x ๏ฝ xf ๏ญ xi ๏ฝ 100 ๏ญ 0 m ๏ฝ 100 m
3. (b) Add the lengths:
15 ๏ซ 100 ๏ซ 30 ๏ซ 100 ๏ซ 15 m ๏ฝ 260 m
4. Subtract xi from xf to find the displacement.
๏ x ๏ฝ xf ๏ญ xi ๏ฝ 0 ๏ญ 0 m ๏ฝ 0 m
Insight: The distance traveled is always positive, but the displacement can be negative. The displacement is always
zero for a complete circuit, as in this case.
Copyright ยฉ 2017 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No
portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
2โ2
James S. Walker, Physics, 5th Edition
Chapter 2: One-Dimensional Kinematics
6.
Picture the Problem: The pony walks around the circular track.
5.25 m
A
Strategy: The distance is the total length of travel, and the
displacement is the net change in position.
B
Solution: (a) 1. The distance traveled is half the circumference:
d ๏ฝ 12 ๏จ 2๏ฐ r ๏ฉ ๏ฝ ๏ฐ r ๏ฝ ๏ฐ ๏จ 5.25 m ๏ฉ ๏ฝ 16.5 m
2. The displacement is the distance from A to B:
๏ x ๏ฝ xf ๏ญ xi ๏ฝ 2r ๏ฝ 2 ๏จ 5.25 m ๏ฉ ๏ฝ 10.5 m
3. (b) The distance traveled will increase when the child completes one circuit, because the pony will have taken more
steps.
4. (c) The displacement will decrease when the child completes one circuit, because the displacement is maximum
when the child has gone halfway around, and is zero when the child returns to the starting position.
d ๏ฝ 2๏ฐ r ๏ฝ 2๏ฐ ๏จ 5.25 m ๏ฉ ๏ฝ 33.0 m
5. (d) The distance traveled equals the circumference:
6. The displacement is zero because the child has returned to her starting position.
Insight: The distance traveled is always positive, but the displacement can be negative. The displacement is always
zero for a complete circuit, as in this case.
7.
Picture the Problem: You drive your car in a straight line at two different speeds.
Strategy: We could calculate the average speed with the given information by determining the total distance traveled
and dividing by the elapsed time. However, we can arrive at a conceptual understanding of the answer by remembering
that average speed is an average over time, not an average over the distance traveled.
Solution: 1. (a) The average speed will be less than 20 m/s because you will spend a longer time driving at the lower
speed. You will cover the second 10 km distance in less time at the higher speed than you did at the lower speed.
2. (b) The best answer is I. More time is spent at 15 m/s than at 25 m/s because the distances traveled at each speed are
the same, so that it will take a longer time at the slower speed to cover the same distance. Statement II is true but
irrelevant and statement III is false.
Insight: The time elapsed at the lower speed is ๏จ10,000 m ๏ฉ ๏จ15 m/s ๏ฉ ๏ฝ 667 s and the time elapsed at the higher speed
is ๏จ10,000 m ๏ฉ ๏จ 25 m/s ๏ฉ ๏ฝ 400 s, hence the average speed is ๏จ 20,000 m ๏ฉ ๏จ1067 s ๏ฉ ๏ฝ 18.7 m/s.
8.
Picture the Problem: You drive your car in a straight line at two different speeds.
Strategy: We could calculate the average speed with the given information by determining the total distance traveled
and dividing by the elapsed time. However, we can arrive at a conceptual understanding of the answer by remembering
that average speed is an average over time, not an average over the distance traveled.
Solution: 1. (a) The average speed will be equal to 20 m/s because you will spend an equal amount of time driving at
the lower speed as at the higher speed. The average speed is therefore the mean value of the two speeds.
2. (b) The best answer is III. Equal time is spent at 15 m/s and 25 m/s because that fact is stated in the question.
Statements I and II are both false.
Insight: The distance traveled at the lower speed would be ๏จ15 m/s ๏ฉ๏จ 600 s ๏ฉ ๏ฝ 9000 m and the distance traveled at the
higher speed would be ๏จ 25 m/s ๏ฉ๏จ 600 s ๏ฉ ๏ฝ 15,000 m so the average speed is ๏จ 24,000 m ๏ฉ ๏จ1200 s ๏ฉ ๏ฝ 20.0 m/s.
Copyright ยฉ 2017 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No
portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
2โ3
James S. Walker, Physics, 5th Edition
Chapter 2: One-Dimensional Kinematics
9.
Picture the Problem: A runner sprints in the forward direction.
Strategy: The average speed is the distance divided by elapsed time.
Solution: Divide the distance by the time:
s๏ฝ
distance 200.0 m
1 mi
3600 s
๏ฝ
๏ฝ 10.42 m/s ๏ด
๏ด
๏ฝ 23.32 mi/h
time
19.19 s
1609 m
1h
Insight: The displacement would be complicated in this case because the 200-m dash usually takes place on a curved
track. Fortunately, the average speed depends upon distance traveled, not displacement.
10. Picture the Problem: A kangaroo hops in the forward direction.
Strategy: The distance is the average speed multiplied by the time elapsed. The time elapsed is the distance divided by
the average speed.
Solution: 1. (a) Multiply the
average speed by the time elapsed:
km ๏ถ๏ฆ
1h ๏ถ
๏ฆ
d ๏ฝ s t ๏ฝ ๏ง 65
๏ท๏ง 3.2 min ๏ด
๏ท ๏ฝ 3.5 km
h
60
min ๏ธ
๏จ
๏ธ๏จ
2. (b) Divide the distance by the average speed:
t๏ฝ
d 0.25 km 60 min
๏ฝ
๏ด
๏ฝ 14 s
s 65 km/h
1h
Insight: The instantaneous speed might vary from 65 km/h, but the time elapsed and the distance traveled depend only
upon the average speed during the interval in question.
11. Picture the Problem: Rubber ducks drift along the ocean surface.
Strategy: The average speed is the distance divided by elapsed time.
Solution: 1. (a) Divide
the distance by the time:
s๏ฝ
d 1600 mi 1609 m 1 mo
1d
๏ฝ
๏ด
๏ด
๏ด
๏ฝ 0.098 m/s
t
10 mo
1 mi
30.5 d 8.64 ๏ด104s
2. (b) Divide the distance by the time:
s๏ฝ
d 1600 mi 1 mo 1 d
๏ฝ
๏ด
๏ด
๏ฝ 0.22 mi/h
t
10 mo 30.5 d 24 h
Insight: The instantaneous speed might vary from 0.098 m/s, but we can calculate only average speed from the total
distance traveled and time elapsed.
12. Picture the Problem: Radio waves propagate in a straight line.
Strategy: The time elapsed is the distance divided by the average speed. The distance to the Moon is 2.39ร105 mi. We
must double this distance because the signal travels there and back again.
t๏ฝ
Solution: Divide the distance by the average speed:
5
2d 2 ๏จ 2.39 ๏ด10 mi ๏ฉ
๏ฝ
๏ฝ 2.57 s
s
1.86 ๏ด105 mi/s
Insight: The time is slightly shorter than this because the given distance is from the center of the Earth to the center of
the Moon, but presumably any radio communications would occur between the surfaces of the Earth and Moon. When
the radii of the two spheres is taken into account, the time decreases to 2.52 s.
13. Picture the Problem: Sound waves propagate in a straight line from a thunderbolt to your ears.
Strategy: The distance is the average speed multiplied by the time elapsed. We will neglect the time it takes for the
light wave to arrive at your eyes because it is vastly smaller than the time it takes the sound wave to travel.
Solution: Multiply the average speed by the time elapsed:
d ๏ฝ s t ๏ฝ ๏จ 340 m/s ๏ฉ๏จ 6.5 s ๏ฉ ๏ฝ 2200 m ๏ฝ 2.2 km
Insight: The speed of sound, 340 m/s, works out to approximately one mile every five seconds, a useful rule of thumb
for estimating the distance to an approaching thunderstorm!
Copyright ยฉ 2017 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No
portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
2โ4
James S. Walker, Physics, 5th Edition
Chapter 2: One-Dimensional Kinematics
14. Picture the Problem: Human nerve impulses propagate at a fixed speed.
Strategy: The time elapsed is the distance divided by the average speed. The distance from your toes to your brain is
on the order of two meters.
t๏ฝ
Solution: Divide the distance by the average speed:
d
2m
๏ฝ
๏ฝ 0.02 s
s 1๏ด102 m/s
Insight: This nerve impulse travel time is not the limiting factor for human reaction time, which is about 0.2 s.
15. Picture the Problem: A finch travels a short distance on the back of the tortoise and a longer distance through the air,
with both displacements along the same direction.
Strategy: First find the total distance traveled by the finch and then determine the average speed by dividing by the
total time elapsed.
Solution: 1. Determine the total distance traveled:
d ๏ฝ s1๏t1 ๏ซ s2 ๏t2
d ๏ฝ ๏ฉ๏ซ๏จ 0.060 m/s ๏ฉ๏จ1.5 min ๏ฉ ๏ซ ๏จ11 m/s ๏ฉ๏จ1.5 min ๏ฉ ๏น๏ป ๏ด 60 s/min
d ๏ฝ 995 m
2. Divide the distance by the time elapsed:
s๏ฝ
d
995 m
๏ฝ
๏ฝ 5.5 m/s
๏t 3.0 min ๏ด 60 s/min
Insight: Most of the distance traveled by the finch occurred by air. In fact, if we neglect the 5.4 m the finch traveled
while on the tortoiseโs back, we still get an average speed of 5.5 m/s over the 3.0 min time interval! The bird might as
well have been at rest during the time it perched on the tortoiseโs back.
16. Picture the Problem: You jog for 5.0 km and then travel an additional 13 km by car, with both displacements along the
same direction.
Strategy: First find the total time elapsed by dividing the distance traveled by the average speed. Find the time elapsed
while jogging, and subtract it from the total time to find the time elapsed while in the car. Finally use the travel-by-car
distance and time information to find the average speed with which you must drive the car.
d
5.0 ๏ซ 13 km
๏ฝ
๏ฝ 0.72 h
sav
25 km/h
Solution: 1. Use the definition of average
speed to determine the total time elapsed.
๏t ๏ฝ
2. Find the time elapsed while jogging:
๏t1 ๏ฝ
3. Find the time elapsed while in the car:
๏t2 ๏ฝ ๏t ๏ญ ๏t1 ๏ฝ 0.72 h ๏ญ 0.55 h ๏ฝ 0.17 h
4. Find the speed of the car:
s2 ๏ฝ
d1
5.0 km
๏ฝ
๏ฝ 0.55 h
v1 9.1 km/h
d 2 13 km
๏ฝ
๏ฝ 76 km/h
๏t2 0.17 h
Insight: Notice that the average speed is not the average of 9.1 km/h and 76 km/h (which would be 43 km/h) because
you spend a much longer time jogging at low speed than you spend driving at high speed.
17. Picture the Problem: A dog continuously runs back and forth as
the owners close the distance between each other.
Strategy: First find the time that will elapse before the owners
meet each other. Then determine the distance the dog will cover
if it continues running at constant speed over that time interval.
2.7 m/s
8.2 m
d
4.10 m
๏ฝ
๏ฝ 3.15 s
sav 1.3 m/s
Solution: 1. Find the time it takes each owner to walk
half the distance (4.10 m) before meeting each other:
๏t ๏ฝ
2. Find the distance the dog runs:
d ๏ฝ s๏t ๏ฝ ๏จ 2.7 m/s ๏ฉ๏จ 3.15 s ๏ฉ ๏ฝ 8.5 m
Insight: The dog will actually run a shorter distance than this, because it is impossible for it to maintain the same
2.7 m/s as it turns around to run to the other owner. It must first slow down to zero speed and then accelerate again.
Copyright ยฉ 2017 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No
portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
2โ5
James S. Walker, Physics, 5th Edition
Chapter 2: One-Dimensional Kinematics
18. Picture the Problem: Blood flows at two different speeds through arteries during a specified time interval.
Strategy: Determine the average speed by first calculating the total distance traveled and then dividing it by the total
time elapsed.
Solution: 1. (a) Because the time intervals are the same, the blood spends equal times at 1.0 m/s and 0.60 m/s, and its
average speed will be equal to 0.80 m/s.
๏จ1.0 m/s ๏ฉ๏จ 0.50 s ๏ฉ ๏ซ ๏จ 0.60 m/s ๏ฉ๏จ 0.50 s ๏ฉ 0.80 m
s ๏t ๏ซ s ๏t
2. (b) Divide the total distance
sav ๏ฝ 1 1 2 2 ๏ฝ
๏ฝ
by the time elapsed:
๏t1 ๏ซ ๏t2
0.50 ๏ซ 0.50 s
1.00 s
๏ฝ 0.80 m/s
Insight: Average speed is a weighted average according to how much time the blood spends traveling at each speed.
19. Picture the Problem: Blood flows at two different speeds through arteries over a specified distance.
Strategy: Determine the average speed by first calculating the total distance traveled and then dividing it by the total
time elapsed.
Solution: 1. (a) The distance intervals are the same but the time intervals are different. The blood will spend more
time at the lower speed than at the higher speed. Because the average speed is a time-weighted average, it will be less
than 0.80 m/s.
2. (b) Divide the total distance by the time elapsed:
sav ๏ฝ
d1 ๏ซ d 2
d ๏ซ d2
1.00 m
๏ฝ 1
๏ฝ
๏t1 ๏ซ ๏t2 d1 d 2 ๏ฆ 0.50 m 0.50 m ๏ถ
๏ซ
๏ซ
๏ง
๏ท
s1 s2
๏จ 1.0 m/s 0.60 m/s ๏ธ
sav ๏ฝ 0.75 m/s
Insight: The blood spends 0.50 s flowing at 1.0 m/s and 0.83 s flowing at 0.60 m/s.
20. Picture the Problem: You travel in a straight line at two different speeds during the specified time interval.
Strategy: Determine the distance traveled during each leg of the trip in order to plot the graph.
Solution: 1. (a) Calculate the
distance traveled in the first leg:
d1 ๏ฝ s1๏t1 ๏ฝ ๏จ12 m/s ๏ฉ๏จ1.5 min ๏ด 60 s/min ๏ฉ ๏ฝ 1080 m
2. Calculate the distance traveled in the second leg:
d2 ๏ฝ s2 ๏t2 ๏ฝ ๏จ 0 m/s ๏ฉ๏จ 3.5 min ๏ฉ ๏ฝ 0 m
3. Calculate the distance traveled in the third leg:
d3 ๏ฝ s3 ๏t3 ๏ฝ ๏จ15 m/s ๏ฉ๏จ 2.5 min ๏ด 60 s/min ๏ฉ ๏ฝ 2250 m
4. Calculate the total distance traveled:
d ๏ฝ d1 ๏ซ d2 ๏ซ d3 ๏ฝ 3330 m
5. Draw the graph:
6. (b) Divide the total distance by the time elapsed:
sav ๏ฝ
d1 ๏ซ d 2 ๏ซ d3
3330 m
๏ฝ
๏ฝ 7.4 m/s
๏t1 ๏ซ ๏t2 ๏ซ ๏t3 7.5 min ๏ด 60 s/min
Insight: The average speed is a weighted average according to how much time you spend traveling at each speed. Here
you spend the most amount of time at rest, so the average speed is less than either 12 m/s or 15 m/s.
Copyright ยฉ 2017 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No
portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
2โ6
James S. Walker, Physics, 5th Edition
Chapter 2: One-Dimensional Kinematics
21. Picture the Problem: As specified in the position-versus-time graph, the father
walks forward, stops, walks forward again, and then walks backward.
Strategy: Determine the direction of the velocity from the slope of the graph along
each segment. Then determine the magnitude of the velocity by calculating the
slope of the graph at each specified point.
Solution: 1. (a) The slope at A is positive so the velocity is positive.
(b) The velocity at B is zero. (c) The velocity at C is positive. (d) The velocity at
D is negative.
๏ x 2.0 m
vav ๏ฝ
๏ฝ
๏ฝ 2.0 m/s
2. (e) Find the slope of the graph at A:
๏t
1.0 s
3. (f) Find the slope of the graph at B:
vav ๏ฝ
๏ x 0.0 m
๏ฝ
๏ฝ 0.0 m/s
๏t 1.0 s
4. (g) Find the slope of the graph at C:
vav ๏ฝ
๏ x 1.0 m
๏ฝ
๏ฝ 1.0 m/s
๏t 1.0 s
5. (h) Find the slope of the graph at D:
vav ๏ฝ
๏ x ๏ญ3.0 m
๏ฝ
๏ฝ ๏ญ1.5 m/s
๏t
2.0 s
Insight: The signs of each answer in (e) through (h) match those predicted in parts (a) through (d). With practice you
can form both a qualitative and quantitative โmovieโ of the motion in your head simply by examining the positionversus-time graph.
22. Picture the Problem: The given position function indicates the particle begins traveling in the negative direction but is
accelerating in the positive direction.
Strategy: Create the x-versus-t plot using a spreadsheet, or calculate individual values by hand and sketch the curve
using graph paper. Use the known x and t information to determine the average velocity. To find the average speed, we
must find the total distance that the particle travels between 0 and 1.0 s, and then divide by 1.0 s.
Solution: 1. (a) Use a spreadsheet or
similar program to create the plot
shown at right. Notice that the average
velocity over the first second of time is
equal to the slope of a straight line
drawn from the origin to the value of
the curve at t = 1.0 s. At that time the
position is โ2.0 m.
2. (b) Find the average velocity
from t = 0 to t = 1.0 s:
3. (c) To find the average speed, we
must determine the distance traveled.
First calculate the time at which x = 0:
2
2
๏ฉ
๏น
๏ x ๏ซ๏จ ๏ญ5 m/s ๏ฉ๏จ1.0 s ๏ฉ ๏ซ ๏จ 3 m/s ๏ฉ ๏จ1.0 s ๏ฉ ๏ป ๏ญ ๏0.0 m๏
vav ๏ฝ
๏ฝ
๏ฝ ๏ญ2.0 m/s
๏t
1.0 s
0 ๏ฝ ๏จ ๏ญ5 m/s ๏ฉ t ๏ซ ๏จ 3 m/s 2 ๏ฉ t 2
5 m/s ๏ฝ ๏จ 3 m/s 2 ๏ฉ t ๏ t ๏ฝ 5 3 s ๏ฝ 1.67 s
4. The time at which the particle turns
around is half the time found in step 3.
Find x at the turnaround time:
x ๏ฝ ๏จ ๏ญ5 m/s ๏ฉ๏จ 5 6 s ๏ฉ ๏ซ ๏จ 3 m/s ๏ฉ๏จ5 6 s ๏ฉ ๏ฝ ๏ญ2.083 m
5. At t = 1 s, the particle is at x = โ2 m, so
it has traveled an additional 0.083 m after
turning around. Find the average speed:
sav ๏ฝ
2
2.083 ๏ซ 0.083 m
๏ฝ 2.2 m/s
1.0 s
Insight: The instantaneous speed is always the magnitude of the instantaneous velocity, but the average speed is not
always the magnitude of the average velocity. For instance, in this problem the particle returns to x = 0 after 1.67 s, at
which time its average speed is sav ๏ฝ 4.17 m 1.67 s ๏ฝ 2.50 m/s, but its average velocity is zero because ๏x = 0.
Copyright ยฉ 2017 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No
portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
2โ7
James S. Walker, Physics, 5th Edition
Chapter 2: One-Dimensional Kinematics
23. Picture the Problem: The given position function indicates the particle begins traveling in the positive direction but is
accelerating in the negative direction.
Strategy: Create the x-versus-t plot using a spreadsheet, or calculate individual values by hand and sketch the curve
using graph paper. Use the known x and t information to determine the average speed and velocity.
Solution: 1. (a) Use a spreadsheet to create the plot shown at right:
๏x
๏t
2. (b) Find the average velocity
from t = 0 to t = 1.0 s:
vav ๏ฝ
3. (c) The average speed is the
magnitude of the average velocity:
sav ๏ฝ vav ๏ฝ 4.0 m/s
๏ฉ๏จ 6 m/s ๏ฉ๏จ1.0 s ๏ฉ ๏ซ ๏จ ๏ญ2 m/s 2 ๏ฉ ๏จ1.0 s ๏ฉ2 ๏น ๏ญ ๏0.0 m ๏
๏ป
๏ฝ๏ซ
1.0 s
vav ๏ฝ 4.0 m/s
Insight: Notice that the average velocity over the first second of time is equal to the slope of a straight line drawn from
the origin to the curve at t = 1.0 s. At that time the position is 4.0 m.
24. Picture the Problem: Following the motion specified in the positionversus-time graph, the tennis player moves left, then right, then left again,
if we take left to be in the negative direction.
Strategy: Determine the direction of the velocity from the slope of the
graph. The speed will be greatest for the segment of the curve that has the
largest slope magnitude.
Solution: 1. (a) The magnitude of the slope at B is larger than A or C so
we conclude the speed is greatest at B.
2. (b) Find the slope of the graph at A:
sav ๏ฝ
3. (c) Find the slope of the graph at B:
sav ๏ฝ
4. (d) Find the slope of the graph at C:
sav ๏ฝ
๏x
๏t
๏x
๏t
๏x
๏t
๏ฝ
๏ฝ
๏ฝ
๏ญ2.0 m
2.0 s
2.0 m
1.0 s
๏ฝ 2.0 m/s
๏ญ1.0 m
2.0 s
๏ฝ 1.0 m/s
๏ฝ 0.50 m/s
Insight: The speed during segment B is larger than the speed during segments A and C, as predicted. Speeds are
always positive because they do not involve direction, but velocities can be negative to indicate their direction.
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portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
2โ8
James S. Walker, Physics, 5th Edition
Chapter 2: One-Dimensional Kinematics
25. Picture the Problem: You travel in the forward direction along the roads leading to the wedding ceremony, but your
average speed is different during the first and second portions of the trip.
Strategy: First find the distance traveled during the first 15 minutes in order to calculate the distance yet to travel.
Then determine the speed you need during the second 15 minutes of travel.
Solution: 1. Use the definition of average
speed to determine the distance traveled:
mi ๏ถ๏ฆ
1h ๏ถ
๏ฆ
d1 ๏ฝ s1๏t1 ๏ฝ ๏ง 5.0
๏ท๏ง15.0 min ๏ด
๏ท ๏ฝ 1.25 mi
h ๏ธ๏จ
60 min ๏ธ
๏จ
2. Find the remaining distance to travel:
d2 ๏ฝ dtotal ๏ญ d1 ๏ฝ 10.0 ๏ญ 1.25 mi ๏ฝ 8.8 mi
3. Find the required speed for
the second part of the trip:
s2 ๏ฝ
d2
8.8 mi
๏ฝ
๏ฝ 35 mi/h
๏t2 0.250 h
Insight: The car needs an average speed of 10 mi/0.5 h = 20 mi/h for the entire trip. However, in order to make it on
time it must go seven times faster in the second half (time-wise) of the trip than it did in the first half of the trip.
26. Picture the Problem: The graph in the problem statement depicts the position of a boat as a
function of time.
Strategy: The velocity of the boat is equal to the slope of its position-versus-time graph.
Solution: By examining the graph we can see that the steepest slope in the negative
direction (down and to the right) is at point C. Therefore, the boat had its most negative
velocity at that time. Points A, B, D, and F all correspond to times of zero velocity because the slope of the graph is
zero at those points. Point E has a large positive slope and we conclude the boat had its most positive velocity at that
time. Therefore, the ranking is: C < A = B = D = F < E.
Insight: The portion of the graph to the left of point B also corresponds to a time of high positive velocity.
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portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
2โ9
James S. Walker, Physics, 5th Edition
Chapter 2: One-Dimensional Kinematics
27. Picture the Problem: The given position function indicates the particle begins traveling in the positive direction but is
accelerating in the negative direction.
Strategy: Create the x-versus-t plot using a spreadsheet, or calculate individual values by hand and sketch the curve
using graph paper. Use the known x and t information to determine the average speed and velocity.
Solution: 1. (a) Use a spreadsheet to create
the plot:
2. (b) Find the average
velocity from t = 0.35 to
t = 0.45 s:
3. (c) Find the average
velocity from t = 0.39 to
t = 0.41 s:
3
3
3
3
๏ฉ
๏น ๏ฉ
๏น
๏x ๏ซ๏จ 2 m/s ๏ฉ๏จ 0.45 s ๏ฉ ๏ญ ๏จ 3 m/s ๏ฉ ๏จ 0.45 s ๏ฉ ๏ป ๏ญ ๏ซ๏จ 2 m/s ๏ฉ๏จ 0.35 s ๏ฉ ๏ญ ๏จ 3 m/s ๏ฉ ๏จ 0.35 s ๏ฉ ๏ป
vav ๏ฝ
๏ฝ
๏t
0.10 s
๏ฝ 0.55 m/s
3
3
3
3
๏ฉ
๏น ๏ฉ
๏น
๏x ๏ซ๏จ 2 m/s ๏ฉ๏จ 0.41 s ๏ฉ ๏ญ ๏จ 3 m/s ๏ฉ ๏จ 0.41 s ๏ฉ ๏ป ๏ญ ๏ซ๏จ 2 m/s ๏ฉ๏จ 0.39 s ๏ฉ ๏ญ ๏จ 3 m/s ๏ฉ ๏จ 0.39 s ๏ฉ ๏ป
vav ๏ฝ
๏ฝ
๏t
0.41 ๏ญ 0.39 s
๏ฝ 0.56 m/s
4. (d) The instantaneous speed at t = 0.40 s will be closer to 0.56 m/s. As the time interval becomes smaller the average
velocity is approaching 0.56 m/s, so we conclude the average speed over an infinitesimally small time interval will be
very close to that value.
Insight: Notice that the instantaneous velocity at 0.40 s is equal to the slope of a straight line drawn tangent to the curve
at that point. Because it is difficult to accurately draw a tangent line, we often resort to mathematical methods like those
illustrated above to determine the instantaneous velocity.
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portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
2 โ 10
James S. Walker, Physics, 5th Edition
Chapter 2: One-Dimensional Kinematics
28. Picture the Problem: The given position function indicates the particle begins traveling in the negative direction but is
accelerating in the positive direction.
Strategy: Create the x-versus-t plot using a spreadsheet, or calculate individual values by hand and sketch the curve
using graph paper. Use the known x and t information to determine the average speed and velocity.
Solution: 1. (a) Use a spreadsheet to create the plot:
2. (b) Find the average velocity
from t = 0.150 to t = 0.250 s:
๏ฆ ๏ฉ๏จ ๏ญ2 m/s ๏ฉ๏จ 0.250 s ๏ฉ ๏ซ ๏จ 3 m/s3 ๏ฉ ๏จ 0.250 s ๏ฉ3 ๏น ๏ญ ๏ถ
๏ป
๏ง๏ซ
๏ท
๏ง
3 ๏ท
3
๏ฉ
๏น
๏ญ2 m/s ๏ฉ๏จ 0.150 s ๏ฉ ๏ซ ๏จ 3 m/s ๏ฉ ๏จ 0.150 s ๏ฉ ๏ท
๏x ๏ง๏จ
๏ซ๏จ
๏ป๏ธ
vav ๏ฝ
๏ฝ
๏ฝ ๏ญ1.63 m/s
๏t
0.250 ๏ญ 0.150 s
3. (c) Find the average velocity
from t = 0.190 to t = 0.210 s:
๏ฆ ๏ฉ๏จ ๏ญ2 m/s ๏ฉ๏จ 0.210 s ๏ฉ ๏ซ ๏จ 3 m/s3 ๏ฉ ๏จ 0.210 s ๏ฉ3 ๏น ๏ญ ๏ถ
๏ป
๏ง๏ซ
๏ท
๏ง
3
๏ฉ๏จ ๏ญ2 m/s ๏ฉ๏จ 0.190 s ๏ฉ ๏ซ ๏จ 3 m/s ๏ฉ ๏จ 0.190 s ๏ฉ3 ๏น ๏ท๏ท
๏ง
๏x ๏จ
๏ซ
๏ป๏ธ
vav ๏ฝ
๏ฝ
๏ฝ ๏ญ1.64 m/s
๏t
0.210 ๏ญ 0.190 s
4. (d) The instantaneous speed at t = 0.200 s will be closer to โ1.64 m/s. As the time interval becomes smaller the
average velocity approaches โ1.64 m/s, and we conclude the average speed over an infinitesimally small time interval
will be very close to that value.
Insight: Notice that the instantaneous velocity at 0.200 s is equal to the slope of a straight line drawn tangent to the
curve at that point. Because it is difficult to accurately draw a tangent line, we often resort to mathematical methods
like those illustrated above to determine the instantaneous velocity.
29. Picture the Problem: You accelerate your car from rest along two on-ramps of different lengths.
Strategy: Use the definitions of average speed and acceleration to compare your motion along the two on-ramps.
Solution: 1. (a) We can reason that because you accelerate between the same initial and final velocities, you must have
the same average speed along both on-ramps. If you have the same average speed, then you will accelerate for a shorter
period of time along the shorter on-ramp A. Your acceleration must be greater to achieve the same final velocity in a
shorter time. We conclude that your acceleration along on-ramp A is greater than your acceleration along on-ramp B.
2. (b) As discussed above, the best explanation is I. The shorter acceleration distance along ramp A requires a greater
acceleration. Statement II is true but is not a complete explanation, and statement III is false.
Insight: We could also set v0 ๏ฝ 0 in the equation, v 2 ๏ฝ v02 ๏ซ 2a๏ x and solve for a: a ๏ฝ v 2 2๏ x From this expression
we can see that for the same final velocity v, you will have a smaller acceleration when you accelerate over the greater
distance ๏ x.
30. Picture the Problem: An airplane accelerates uniformly along a straight runway.
Strategy: The average acceleration is the change of the velocity divided by the elapsed time.
๏v 156 ๏ญ 0 mi/h 0.447 m/s
๏ฝ
๏ด
๏ฝ 1.98 m/s2
๏t
35.2 s
mi/h
Insight: The instantaneous acceleration might vary from 1.98 m/s2, but we can calculate only average acceleration from
the net change in velocity and time elapsed.
aav ๏ฝ
Solution: Divide the change in velocity by the time:
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portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
2 โ 11
James S. Walker, Physics, 5th Edition
Chapter 2: One-Dimensional Kinematics
31. Picture the Problem: A runner accelerates uniformly along a straight track.
Strategy: The change in velocity is the average acceleration multiplied by the elapsed time.
Solution: 1. (a) Multiply the acceleration by the time:
v ๏ฝ v0 ๏ซ a t ๏ฝ 0 m/s ๏ซ ๏จ1.9 m/s2 ๏ฉ ๏จ 2.0 s ๏ฉ ๏ฝ 3.8 m/s
2. (b) The runnerโs speed will be the same at the end
of the race as it is at t = 5.2 s:
v ๏ฝ v0 ๏ซ a t ๏ฝ 0 m/s ๏ซ ๏จ1.9 m/s2 ๏ฉ ๏จ 5.2 s ๏ฉ ๏ฝ 9.9 m/s
Insight: World class sprinters have top speeds of over 10 m/s and can get up to speed in much less than 5.2 s.
32. Picture the Problem: An airplane slows down uniformly along a straight runway as it travels toward the east.
Strategy: The average acceleration is the change of the velocity divided by the elapsed time. Assume that east is in the
positive direction.
Solution: 1. Divide the change in velocity by the time:
aav ๏ฝ
vf ๏ญ vi 0 ๏ญ 70.6 m/s
๏ฝ
๏ฝ ๏ญ5.43 m/s 2
๏t
13.0 s
2. We note from the previous step that the acceleration is negative. Because east is the positive direction, negative
acceleration must be toward the west. Thus the jet has an acceleration of 5.43 m/s2 toward the west.
Insight: In physics we almost never talk about deceleration. Instead, we call it negative acceleration.
33. Picture the Problem: A car travels in a straight line due north, either speeding up or slowing down, depending upon the
direction of the acceleration.
Strategy: Use the definition of acceleration to determine the final velocity over the specified time interval. Let north be
the positive direction.
Solution: 1. (a) Calculate the velocity:
v ๏ฝ v0 ๏ซ at ๏ฝ 23.6 m/s ๏ซ ๏จ1.30 m/s2 ๏ฉ ๏จ 7.10 s ๏ฉ ๏ฝ 32.8 m/s north
2. (b) Calculate the velocity:
v ๏ฝ v0 ๏ซ at ๏ฝ 23.6 m/s ๏ซ ๏จ ๏ญ1.15 m/s2 ๏ฉ ๏จ 7.10 s ๏ฉ ๏ฝ 15.4 m/s north
Insight: In physics we almost never talk about deceleration. Instead, we call it negative acceleration. In this problem
south is considered the negative direction, and in part (b) the car is slowing down or undergoing negative acceleration.
34. Picture the Problem: Following the motion specified in the velocityversus-time graph, the motorcycle is speeding up, then moving at constant
speed, then slowing down.
Strategy: Determine the acceleration from the slope of the graph.
Solution: 1. (a) Find the slope at A:
aav ๏ฝ
๏ v 10 m/s
๏ฝ
๏t
5.0 s
๏ฝ 2.0 m/s 2
2. (b) Find the slope of the graph at B:
aav ๏ฝ
๏ v 0 m/s
๏ฝ
๏ฝ 0.0 m/s 2
๏t 10.0 s
3. (c) Find the slope of the graph at C:
aav ๏ฝ
๏ v ๏ญ5.0 m/s
๏ฝ
๏ฝ ๏ญ 0.50 m/s 2
๏t
10.0 s
Insight: The acceleration during segment A is larger than the acceleration during segments B and C because the slope
there has the greatest magnitude.
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portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
2 โ 12
James S. Walker, Physics, 5th Edition
Chapter 2: One-Dimensional Kinematics
35. Picture the Problem: Following the motion specified in the velocityversus-time graph, the person on horseback is speeding up, then
accelerating at an even greater rate, then slowing down.
Strategy: We could determine the acceleration from the slope of the
graph, and then use the acceleration and initial velocity to determine the
displacement. Alternatively, we could use the initial and final velocities
in each segment to determine the average velocity and the time elapsed to
find the displacement during each interval.
Solution: 1. (a) Use the average velocity during
interval A to calculate the displacement:
๏ x ๏ฝ 12 ๏จ v0 ๏ซ v ๏ฉ t ๏ฝ 12 ๏จ 0 ๏ซ 2.0 m/s ๏ฉ๏จ10 s ๏ฉ ๏ฝ 10 m
2. (b) Calculate the displacement during segment B:
๏ x ๏ฝ 12 ๏จ v0 ๏ซ v ๏ฉ t ๏ฝ 12 ๏จ 2.0 ๏ซ 6.0 m/s ๏ฉ๏จ 5.0 s ๏ฉ ๏ฝ 20 m
3. (c) Calculate the displacement during segment C:
๏ x ๏ฝ 12 ๏จ v0 ๏ซ v ๏ฉ t ๏ฝ 12 ๏จ 6.0 ๏ซ 2.0 m/s ๏ฉ๏จ10 s ๏ฉ ๏ฝ 40 m
Insight: There are often several ways to solve motion problems involving constant acceleration, some easier than
others.
36. Picture the Problem: A horse travels in a straight line in the positive direction while accelerating in the negative
direction (slowing down).
Strategy: Use the definition of acceleration to determine the time elapsed for the specified change in velocity.
Solution: Calculate the time interval:
t๏ฝ
v ๏ญ v0 5.5 ๏ญ 9.2 m/s
๏ฝ
๏ฝ 2.0 s
a
๏ญ1.81 m/s 2
Insight: An acceleration of greater magnitude would decrease the horseโs velocity in a shorter period of time.
37. Picture the Problem: Your car travels in a straight line in the positive direction while accelerating in the negative
direction (slowing down).
Strategy: Use the constant acceleration equation of motion to determine the time elapsed for the specified change in
velocity.
Solution: 1. (a) The time required to come to a stop is the change in velocity divided by the acceleration. In both cases
the final velocity is zero, so the change in velocity doubles when you double the initial velocity. Therefore, the stopping
time will increase by a factor of two when you double your driving speed.
2. (b) Calculate the stopping time:
t๏ฝ
v ๏ญ v0 0 ๏ญ 18 m/s
๏ฝ
๏ฝ 4.3 s
a
๏ญ 4.2 m/s 2
3. (c) Calculate the stopping time:
t๏ฝ
v ๏ญ v0 0 ๏ญ 36 m/s
๏ฝ
๏ฝ 8.6 s
a
๏ญ 4.2 m/s 2
Insight: Notice that the deceleration is treated as a negative acceleration in this problem and elsewhere in the text.
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portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
2 โ 13
James S. Walker, Physics, 5th Edition
Chapter 2: One-Dimensional Kinematics
38. Picture the Problem: A train travels in a straight line in the positive direction while accelerating in the positive
direction (speeding up).
Strategy: First find the acceleration and then determine the final velocity.
a๏ฝ
Solution: 1. Use the definition of acceleration:
v ๏ญ v0 4.7 ๏ญ 0 m/s
๏ฝ
๏ฝ 0.94 m/s 2
t
5.0 s
v ๏ฝ v0 ๏ซ a t ๏ฝ 4.7 m/s ๏ซ ๏จ 0.94 m/s 2 ๏ฉ ๏จ 5.0 s ๏ฉ
2. Calculate the final speed of the second segment, using
the final speed from the first segment as the initial speed:
v ๏ฝ 9.4 m/s
Insight: Another way to tackle this problem is to set up similar triangles on a velocity-versus-time graph. The answer
would then be calculated as v = (4.7 m/s) ร 10 s / 5 s = 9.4 m/s. Try it!
39. Picture the Problem: A particle travels in a straight line in the positive direction while accelerating in the positive
direction (speeding up).
Strategy: Use the constant acceleration equation of motion to find the initial velocity.
Solution: Calculate v0 :
v0 ๏ฝ v ๏ญ at ๏ฝ 9.31 m/s ๏ญ ๏จ 6.24 m/s2 ๏ฉ ๏จ 0.450 s ๏ฉ ๏ฝ 6.50 m/s
Insight: As expected, the initial velocity is less than the final velocity because the particle is speeding up.
40. Picture the Problem: A jet travels in a straight line toward the south while accelerating in the northerly direction
(slowing down).
Strategy: Use the relationship between acceleration, velocity, and displacement (Equation 2-12). The acceleration
should be negative if we take the direction of the jetโs motion (to the south) to be positive.
v 2 ๏ญ v02 0 ๏ญ ๏จ 71.4 m/s ๏ฉ
๏ฝ
๏ฝ ๏ญ2.69 m/s 2 ๏ฝ 2.69 m/s2 to the north
2๏ x
2 ๏จ 949 m ๏ฉ
2
Solution: Solve for the acceleration:
a๏ฝ
2
Insight: The negative acceleration indicates the jet is slowing down during that time interval. Notice that Equation 2-12
is a good choice for problems in which no time information is given or requested.
41. Picture the Problem: Your car travels in a straight line toward the west while accelerating in the easterly direction
(slowing down).
Strategy: The average velocity is simply half the sum of the initial and final velocities because the acceleration is
uniform.
Solution: Calculate half the sum of the velocities:
vav ๏ฝ 12 ๏จ v0 ๏ซ v ๏ฉ ๏ฝ 12 ๏จ18 ๏ซ 0 m/s ๏ฉ ๏ฝ 9.0 m/s to the west
Insight: The average velocity of any object that slows down and comes to a stop is just half the initial velocity.
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portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
2 โ 14
James S. Walker, Physics, 5th Edition
Chapter 2: One-Dimensional Kinematics
42. Picture the Problem: A ball rolls down an inclined plane with constant acceleration.
Strategy: The ball starts at a positive value of its position x and must therefore travel in the negative direction in order
to reach the location x = 0.
Solution: 1. (a) No matter how fast the ball might initially move in the positive direction, away from x = 0, a constant
negative acceleration will eventually slow it down, bring it briefly to rest, and speed it up back toward x = 0. Therefore,
in cases 3 and 4, where a < 0, the ball will certainly pass x = 0.
2. (b) It is possible for the initial velocity to be so large in the negative direction that a positive acceleration cannot bring
it to rest before it passes x = 0. Therefore, in case 2 where v0 ๏ผ 0 and a ๏พ 0, it is possible that the ball will pass x = 0,
but we need more information about the relative magnitudes of v0 and a in order to be certain.
3. (c) Whenever the initial velocity is opposite in sign to the acceleration, the ball will eventually come to rest briefly
and then speed up in the direction of the acceleration. Therefore, in cases 2 and 3 we know that the ball will
momentarily come to rest.
Insight: If we suppose that a = +4.00 m/s2 and that x0 ๏ฝ 2.00 m, we can determine that an initial velocity of
v02 ๏ฝ v2 ๏ญ 2 a ๏ x ๏ฝ 02 ๏ญ 2 ๏จ 4.00 m/s2 ๏ฉ ๏จ ๏ญ2.00 m ๏ฉ ๏ v0 ๏ฝ ๏ญ 8 ๏ฝ ๏ญ2.83 m/s is the threshold initial velocity for the ball
to reach the x = 0 position. With that initial velocity the ball will come to rest momentarily at x = 0 before speeding up
in the positive direction again.
43. Picture the Problem: A boat travels in a straight line with constant positive acceleration.
Strategy: The average speed is simply half the sum of the initial and final velocities because the acceleration is
uniform.
Solution: 1. (a) Calculate half the sum of the velocities:
vav ๏ฝ 12 ๏จ v0 ๏ซ v ๏ฉ ๏ฝ 12 ๏จ 0 ๏ซ 4.82 m/s ๏ฉ ๏ฝ 2.41 m/s
2. (b) The distance traveled is the average
velocity multiplied by the time elapsed:
d ๏ฝ vav t ๏ฝ ๏จ 2.41 m/s ๏ฉ๏จ 4.77 s ๏ฉ ๏ฝ 11.5 m
Insight: The average velocity of any object that speeds up from rest is just half the final velocity.
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portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
2 โ 15
James S. Walker, Physics, 5th Edition
Chapter 2: One-Dimensional Kinematics
44. Picture the Problem: The given position function indicates the car begins at a positive position, but is traveling in the
negative direction and accelerating in the negative direction.
Strategy: Compare the given position as a function of time with the symbolic expression to determine the initial
position, initial velocity, and acceleration of the car. Create the x-versus-t plot using a spreadsheet, or calculate
individual values by hand and sketch the curve using graph paper. Finally, use the known x and t information to
determine the distance traveled and the average velocity.
Solution: 1. (a) Compare the
symbolic formula with the given
equation to find the initial position:
2. Compare the symbolic formula
with the given equation to find the
initial velocity:
3. Compare the symbolic formula
with the given equation to find
acceleration:
x ๏ฝ x0 ๏ซ v0 t ๏ซ 12 a t 2 ๏ฝ ๏จ 50 m ๏ฉ ๏ซ ๏จ ๏ญ5.0 m/s ๏ฉ t ๏ซ ๏จ ๏ญ10 m/s 2 ๏ฉ t 2
๏ x0 ๏ฝ 50 m
x ๏ฝ x0 ๏ซ v0 t ๏ซ 12 a t 2 ๏ฝ ๏จ 50 m ๏ฉ ๏ซ ๏จ ๏ญ5.0 m/s ๏ฉ t ๏ซ ๏จ ๏ญ10 m/s 2 ๏ฉ t 2
๏ v0 ๏ฝ ๏ญ5.0 m/s
x ๏ฝ x0 ๏ซ v0 t ๏ซ 12 a t 2 ๏ฝ ๏จ 50 m ๏ฉ ๏ซ ๏จ ๏ญ5.0 m/s ๏ฉ t ๏ซ ๏จ ๏ญ10 m/s 2 ๏ฉ t 2
1
2
a ๏ฝ ๏ญ10 m/s 2
๏ a ๏ฝ ๏ญ20 m/s 2
4. (b) Use a spreadsheet or similar
program to create the blue plot shown at
right. The average velocity of the car
between 1.0 and 2.0 s is equal to the slope
of a straight line drawn from its position at
t = 1.0 s and that at t =2.0 s as shown.
5. (c) Because the car travels in a
straight line and does not reverse
direction, the distance traveled equals
the magnitude of the displacement:
6. (d) Find the average velocity
from t = 1.0 s to t = 2.0 s:
xi ๏ฝ ๏จ 50 m ๏ฉ ๏ซ ๏จ ๏ญ5.0 m/s ๏ฉ๏จ 0 s ๏ฉ ๏ซ ๏จ ๏ญ10 m/s 2 ๏ฉ ๏จ 0 s ๏ฉ ๏ฝ 50 m
2
xf ๏ฝ ๏จ 50 m ๏ฉ ๏ซ ๏จ ๏ญ5.0 m/s ๏ฉ๏จ1.0 s ๏ฉ ๏ซ ๏จ ๏ญ10 m/s 2 ๏ฉ ๏จ1.0 s ๏ฉ ๏ฝ 35 m
2
๏ x ๏ฝ xf ๏ญ xi ๏ฝ 35 ๏ญ 50 m ๏ฝ ๏ญ15 m ๏ distance ๏ฝ ๏ x ๏ฝ 15 m
xi ๏ฝ ๏จ 50 m ๏ฉ ๏ซ ๏จ ๏ญ5.0 m/s ๏ฉ๏จ1.0 s ๏ฉ ๏ซ ๏จ ๏ญ10 m/s 2 ๏ฉ ๏จ1.0 s ๏ฉ ๏ฝ 35 m
2
xf ๏ฝ ๏จ 50 m ๏ฉ ๏ซ ๏จ ๏ญ5.0 m/s ๏ฉ๏จ 2.0 s ๏ฉ ๏ซ ๏จ ๏ญ10 m/s 2 ๏ฉ ๏จ 2.0 s ๏ฉ ๏ฝ 0 m
2
vav ๏ฝ
๏ x xf ๏ญ xi
0 ๏ญ 35 m
๏ฝ
๏ฝ
๏ฝ ๏ญ35 m/s
๏t
t f ๏ญ ti
2.0 ๏ญ 1.0 s
Insight: Average speed and average velocity are always the same as long as the object continuously travels in the same
direction. If it reverses course or travels in two (or three) dimensions, the relationship between the two is more
complex, but the distance traveled will always be greater than or equal to the displacement, so the average speed will
always be greater than or equal to the average velocity.
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portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
2 โ 16
James S. Walker, Physics, 5th Edition
Chapter 2: One-Dimensional Kinematics
45. Picture the Problem: The given position function indicates the ball begins traveling in the positive direction but is
accelerating in the negative direction.
Strategy: Compare the given position as a function of time with the symbolic expression to determine the initial
position, initial velocity, and acceleration of the ball. Create the x-versus-t plot using a spreadsheet, or calculate
individual values by hand and sketch the curve using graph paper. Finally, use the known x and t information to
determine the average velocity and the average speed.
Solution: 1. (a) Compare the
symbolic formula with the given
equation to find the initial position:
2. Compare the symbolic formula
with the given equation to find the
initial velocity:
3. Compare the symbolic formula
with the given equation to find
acceleration:
x ๏ฝ x0 ๏ซ v0 t ๏ซ 12 a t 2 ๏ฝ ๏จ 0 m ๏ฉ ๏ซ ๏จ 5.0 m/s ๏ฉ t ๏ซ ๏จ ๏ญ10 m/s 2 ๏ฉ t 2
๏ x0 ๏ฝ 0 m
x ๏ฝ x0 ๏ซ v0 t ๏ซ 12 a t 2 ๏ฝ ๏จ 0 m ๏ฉ ๏ซ ๏จ 5.0 m/s ๏ฉ t ๏ซ ๏จ ๏ญ10 m/s 2 ๏ฉ t 2
๏ v0 ๏ฝ 5.0 m/s
x ๏ฝ x0 ๏ซ v0 t ๏ซ 12 a t 2 ๏ฝ ๏จ 0 m ๏ฉ ๏ซ ๏จ 5.0 m/s ๏ฉ t ๏ซ ๏จ ๏ญ10 m/s 2 ๏ฉ t 2
1
2
a ๏ฝ ๏ญ10 m/s 2
๏ a ๏ฝ ๏ญ20 m/s 2
4. (b) Use a spreadsheet or similar
program to create the blue plot shown at
right. The average speed of the ball
between 1.0 and 2.0 s is equal to the slope
of a straight line drawn from its position at
t = 1.0 s and that at t =2.0 s as shown.
5. (c) Because the ball reverses
direction, the average velocity from
t = 0 to t = 1.0 s should be calculated
with careful attention to the signs:
xi ๏ฝ ๏จ 0 m ๏ฉ ๏ซ ๏จ 5.0 m/s ๏ฉ๏จ 0 s ๏ฉ ๏ซ ๏จ ๏ญ10 m/s 2 ๏ฉ ๏จ 0 s ๏ฉ ๏ฝ 0 m
2
xf ๏ฝ ๏จ 0 m ๏ฉ ๏ซ ๏จ 5.0 m/s ๏ฉ๏จ1.0 s ๏ฉ ๏ซ ๏จ ๏ญ10 m/s 2 ๏ฉ ๏จ1.0 s ๏ฉ ๏ฝ ๏ญ5.0 m
2
๏ x ๏ฝ xf ๏ญ xi ๏ฝ ๏ญ5.0 ๏ญ 0 m ๏ฝ ๏ญ5 m
vav ๏ฝ ๏ x ๏ t ๏ฝ ๏จ ๏ญ5.0 m ๏ฉ ๏จ1.0 s ๏ฉ ๏ฝ ๏ญ5.0 m/s
6. (d) Because the ball does not
reverse direction between t = 1.0 s to
t = 2.0 s, the average speed is the
magnitude of the average velocity:
xi ๏ฝ ๏จ 0 m ๏ฉ ๏ซ ๏จ 5.0 m/s ๏ฉ๏จ1.0 s ๏ฉ ๏ซ ๏จ ๏ญ10 m/s 2 ๏ฉ ๏จ1.0 s ๏ฉ ๏ฝ ๏ญ5.0 m
2
xf ๏ฝ ๏จ 0 m ๏ฉ ๏ซ ๏จ 5.0 m/s ๏ฉ๏จ 2.0 s ๏ฉ ๏ซ ๏จ ๏ญ10 m/s 2 ๏ฉ ๏จ 2.0 s ๏ฉ ๏ฝ ๏ญ30 m
2
vav ๏ฝ
๏ x xf ๏ญ xi ๏ญ30 ๏ญ ๏จ ๏ญ5.0 ๏ฉ m
๏ฝ
๏ฝ
๏ฝ ๏ญ25 m/s ๏ vav ๏ฝ 25 m/s
๏t
t f ๏ญ ti
2.0 ๏ญ 1.0 s
Insight: The instantaneous speed is always the magnitude of the instantaneous velocity, but the average speed is not
always the magnitude of the average velocity. For instance, in this problem the ball travels to +0.625 m at t = 0.250 s
and then to โ5.00 m at t = 1.00 s, a total distance of 6.25 m, while its displacement is โ5.00 m. Hence its average speed
is 6.25 m/s while its average velocity is โ5.00 m/s over the time interval between t = 0 and t = 1.0 s
Copyright ยฉ 2017 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No
portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
2 โ 17
James S. Walker, Physics, 5th Edition
Chapter 2: One-Dimensional Kinematics
46. Picture the Problem: A cheetah runs in a straight line with constant positive acceleration.
Strategy: The average velocity is simply half the sum of the initial and final velocities because the acceleration is
uniform. The distance traveled is the average velocity multiplied by the time elapsed.
Solution: 1. (a) Calculate half the sum of the velocities:
vav ๏ฝ 12 ๏จ v0 ๏ซ v ๏ฉ ๏ฝ 12 ๏จ 0 ๏ซ 25.0 m/s ๏ฉ ๏ฝ 12.5 m/s
2. Use the average velocity to find the distance:
d ๏ฝ vavt ๏ฝ ๏จ12.5 m/s ๏ฉ๏จ 6.22 s ๏ฉ ๏ฝ 77.8 m
3. (b) For a constant acceleration the velocity varies linearly with time. Therefore we expect the velocity to be equal to
12.5 m/s after half the time (3.11 s) has elapsed.
4. (c) Calculate half the sum of the velocities:
vav,1 ๏ฝ 12 ๏จ v0 ๏ซ v ๏ฉ ๏ฝ 12 ๏จ 0 ๏ซ 12.5 m/s ๏ฉ ๏ฝ 6.25 m/s
5. Calculate half the sum of the velocities:
vav,2 ๏ฝ 12 ๏จ v0 ๏ซ v ๏ฉ ๏ฝ 12 ๏จ12.5 ๏ซ 25.0 m/s ๏ฉ ๏ฝ 18.8 m/s
6. (d) Use the average velocity to find the distance:
d1 ๏ฝ vav,1 t ๏ฝ ๏จ 6.25 m/s ๏ฉ๏จ 3.11 s ๏ฉ ๏ฝ 19.4 m
7. Use the average velocity to find the distance:
d2 ๏ฝ vav,2 t ๏ฝ ๏จ18.8 m/s ๏ฉ๏จ 3.11 s ๏ฉ ๏ฝ 58.5 m
Insight: The distance traveled is always the average velocity multiplied by the time. This is a consequence of the
definition of average velocity.
Copyright ยฉ 2017 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No
portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
2 โ 18
James S. Walker, Physics, 5th Edition
Chapter 2: One-Dimensional Kinematics
47. Picture the Problem: Measurements taken from a video of a sled traveling down an icy slope can be used to determine
the average speed and the acceleration of the sled.
Strategy: Create an x-versus-t plot using a spreadsheet, or plot the individual values by hand using graph paper. Use
the known x and t information to determine the average velocity over the specified time interval. Use either the
spreadsheet features or the equation x ๏ฝ x0 ๏ซ v0 t ๏ซ 12 a t 2 to determine the average acceleration of the sled.
Solution: 1. (a) Use a spreadsheet or
similar program to create the plot shown at
right. The average velocity of the sled
between 0.25 and 1.3 s is equal to the
slope of a straight line drawn from its
position at t =0.25 s and that at t =1.3 s as
shown.
2. (b) Draw a smooth curve to represent
the sled data, and use the smooth curve
to determine the approximate average
speed:
3. Check your answer using the
least-squares regression from
the spreadsheet:
From the plot, xi ๏ป 0.15 m and xf ๏ป 2.5 m
vav ๏ฝ
๏ x 2.5 ๏ญ 0.15 m
๏ฝ
๏ฝ 2.2 m/s
๏t
1.3 ๏ญ 0.25 s
xi ๏ฝ ๏จ ๏ญ 0.0093 m ๏ฉ ๏ซ ๏จ ๏ญ 0.0224 m/s ๏ฉ๏จ 0.25 s ๏ฉ ๏ซ ๏จ1.4243 m/s 2 ๏ฉ ๏จ 0.25 s ๏ฉ
2
๏ฝ 0.074 m
xf ๏ฝ ๏จ ๏ญ 0.0093 m ๏ฉ ๏ซ ๏จ ๏ญ 0.0224 m/s ๏ฉ๏จ1.3 s ๏ฉ ๏ซ ๏จ1.4243 m/s2 ๏ฉ ๏จ1.3 s ๏ฉ
2
๏ฝ 2.37 m
๏ x xf ๏ญ xi 2.37 ๏ญ 0.074 m
vav ๏ฝ
๏ฝ
๏ฝ
๏ฝ 2.19 m/s ๏ confirmed
๏t
t f ๏ญ ti
1.3 ๏ญ 0.25 s
4. (c) Calculate the average
acceleration of the sled from the
equation, x ๏ฝ x0 ๏ซ v0 t ๏ซ 12 a t 2 :
5. Check your answer using the leastsquares regression from the spreadsheet:
x ๏ฝ x0 ๏ซ v0 t ๏ซ 12 a t 2 ๏ฝ 0 ๏ซ 0 ๏ซ 12 a t 2
a๏ฝ
1
2
2 x 2 ๏จ 8.8 m ๏ฉ
๏ฝ
๏ฝ 2.8 m/s 2
2
t2
2.5
s
๏จ
๏ฉ
a ๏ฝ 1.4243 m/s2
๏ a ๏ฝ 2.8486 m/s2 ๏ confirmed
Insight: Spreadsheet software usually includes powerful tools like regression to analyze data like these.
48. Picture the Problem: A child slides down the hill in a straight line with constant positive acceleration.
Strategy: Use the known acceleration and times to determine the positions of the child. In each case x0 and v0 are
zero.
Solution: 1. (a) Calculate her position:
x ๏ฝ x0 ๏ซ v0 t ๏ซ 12 a t 2 ๏ฝ 0 ๏ซ 0 ๏ซ 12 ๏จ1.6 m/s2 ๏ฉ ๏จ1.0 s ๏ฉ ๏ฝ 0.80 m
2. (b) Calculate her position:
x ๏ฝ x0 ๏ซ v0 t ๏ซ 12 a t 2 ๏ฝ 0 ๏ซ 0 ๏ซ 12 ๏จ1.6 m/s2 ๏ฉ ๏จ 2.0 s ๏ฉ ๏ฝ 3.2 m
3. (c) Calculate her position:
x ๏ฝ x0 ๏ซ v0 t ๏ซ 12 a t 2 ๏ฝ 0 ๏ซ 0 ๏ซ 12 ๏จ1.6 m/s2 ๏ฉ ๏จ 3.0 s ๏ฉ ๏ฝ 7.2 m
2
2
2
Insight: Her position varies with the square of the time because of her constant acceleration.
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portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
2 โ 19
James S. Walker, Physics, 5th Edition
Chapter 2: One-Dimensional Kinematics
49. Picture the Problem: Passengers on the Detonator ride accelerate straight downward.
Strategy: Use the known initial and final velocities and the elapsed time to find the acceleration.
a๏ฝ
Solution: Calculate the acceleration:
v f ๏ญ vi
๏t
๏ฝ
๏จ 45 ๏ญ 0 mi/h ๏ฉ
2.2 s
๏ด
0.447 m/s
๏ฝ 9.1 m/s 2
mi/h
Insight: The passengerโs acceleration is just less than that for a free-falling object. What a thrill!
50. Picture the Problem: Jules Verneโs Columbiad spaceship accelerates from rest down the barrel of the cannon.
Strategy: Employ the relationship between acceleration, displacement, and velocity (Equation 2-12) to find the
acceleration.
v 2 ๏ญ v02 ๏จ12000 yd/s ๏ด 3 ft/yd ๏ด 0.305 m/ft ๏ฉ ๏ญ 0
a๏ฝ
๏ฝ
๏ฝ 2.8 ๏ด105 m/s2
2๏x
2 ๏จ 700 ft ๏ด 0.305 m/ft ๏ฉ
2
Solution: Calculate the acceleration:
2
Insight: An acceleration this great would tear the occupants of the spacecraft apart! Notice that the equation
v2 ๏ฝ v02 ๏ซ 2 a ๏ x is a good choice for problems in which no time information is given or requested.
51. Picture the Problem: An Escherichia coli bacterium accelerates from rest in the forward direction.
Strategy: Employ the definition of acceleration to find the time elapsed, and the relationship between acceleration,
displacement, and velocity (Equation 2-12) to find the distance traveled.
v ๏ญ v0 12 ๏ญ 0 ๏ญ m/s
๏ฝ
๏ฝ 0.077 s
a
156 ๏ญ m/s 2
Solution: 1. (a) Calculate the time to accelerate:
t๏ฝ
2. (b) Calculate the displacement:
2
v 2 ๏ญ v02 ๏จ12 ๏ญ m/s ๏ฉ ๏ญ 0
๏x ๏ฝ
๏ฝ
๏ฝ 0.46 ๏ญ m
2a
2 ๏จ156 ๏ญ m/s 2 ๏ฉ
2
Insight: The accelerations are tiny but so are the bacteria! The average speed here is about 3 body lengths per second if
each bacterium were 2 ยตm long. If this were a human that would be 6 m/s or 13 mi/h, much faster than we can swim!
52. Picture the Problem: Two cars are traveling in
opposite directions.
Strategy: Write the equations of motion based upon
Equation 2-11, and set them equal to each other to
find the time at which the two cars pass each other.
Solution: 1. (a) Write an equation for the
position of car 1, which is traveling east and
speeding up. Let east be the positive direction:
x1 ๏ฝ x0,1 ๏ซ v0,1 t ๏ซ 12 a1 t 2 ๏ฝ x1 ๏ฝ ๏จ 20.0 m/s ๏ฉ t ๏ซ 12 ๏จ 2.5 m/s2 ๏ฉ t 2
2. Write an equation for the position of car 2,
which is traveling west but slowing down,
which means it is accelerating toward the east:
x2 ๏ฝ x0,2 ๏ซ v0,2 t ๏ซ 12 a2 t 2 ๏ฝ x2 ๏ฝ 100 m ๏ญ ๏จ 30.0 m/s ๏ฉ t ๏ซ 12 ๏จ 3.2 m/s2 ๏ฉ t 2
3. (b) Set x1 ๏ฝ x2 and solve for t:
๏จ 20.0 m/s ๏ฉ t ๏ซ ๏จ1.25 m/s2 ๏ฉ t 2 ๏ฝ 100 m ๏ญ ๏จ 30.0 m/s ๏ฉ t ๏ซ ๏จ1.6 m/s2 ๏ฉ t 2
0 ๏ฝ 100 ๏ญ 50t ๏ซ 0.35t 2
t๏ฝ
50 ๏ฑ 502 ๏ญ 4 ๏จ 0.35๏ฉ๏จ100 ๏ฉ
0.70
๏ฝ 2.03, 141 s ๏ 2.0 s
Insight: We chose smaller of the two roots, which corresponds to the first time the cars pass each other. The larger
acceleration of car 2 means that itโll come to rest, speed up in the positive direction, and overtake car 1 at t = 141 s.
Copyright ยฉ 2017 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No
portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
2 โ 20
James S. Walker, Physics, 5th Edition
Chapter 2: One-Dimensional Kinematics
53. Picture the Problem: A meteorite accelerates from a high speed to rest after impacting the car.
Strategy: Employ the relationship between acceleration, displacement, and velocity (Equation 2-12) to find the
acceleration.
a ๏ฝ
Solution: Calculate the acceleration:
v 2 ๏ญ v02
2 ๏x
๏ฝ
02 ๏ญ ๏จ130 m/s ๏ฉ
2 ๏จ 0.22 m ๏ฉ
2
๏ฝ 3.8 ๏ด104 m/s 2
Insight: The high stiffness of steel is responsible for the tremendous (negative) acceleration of the meteorite.
54. Picture the Problem: A rocket accelerates straight upward.
Strategy: Employ the relationship between acceleration, displacement, and time (Equation 2-11) to find the
acceleration. Because the rocket was at rest before blast off, the initial velocity v0 is zero, and so is the initial position
x0 . Once the acceleration is known, we can use the constant acceleration equation (Equation 2-7) to find the speed.
Solution: 1. (a) Write out the position vs. time equation:
x ๏ฝ x0 ๏ซ v0 t ๏ซ 12 a t 2
2. Let x0 ๏ฝ v0 ๏ฝ 0 and solve for acceleration:
a๏ฝ
3. (b) Calculate the final speed:
v ๏ฝ 0 ๏ซ a t ๏ฝ ๏จ 23.2 m/s2 ๏ฉ ๏จ 2.8 s ๏ฉ ๏ฝ 65 m/s
2 x 2 ๏จ 91 m ๏ฉ
๏ฝ
๏ฝ 23 m/s 2 upward
2
t2
๏จ 2.8 s ๏ฉ
Insight: The position vs. time equation simplifies considerably if the initial position and the initial velocity are zero.
55. Picture the Problem: You drive in a straight line and then slow down to a stop.
Strategy: Employ the relationship between acceleration, displacement, and velocity (Equation 2-12) to find the
displacement. Equation 2-12 is a good choice for problems in which no time information is given or requested. In this
case the acceleration is negative because the car is slowing down.
v 2 ๏ญ v02 02 ๏ญ v02
v2
๏จ12.0 m/s ๏ฉ
๏ฝ
๏ฝ๏ญ 0 ๏ฝ๏ญ
๏ฝ 21 m
2a
2a
2a
2 ๏จ ๏ญ3.5 m/s 2 ๏ฉ
2
Solution: 1. (a) Calculate the displacement:
๏x ๏ฝ
2. (b) Because velocity is proportional to the square root of displacement, cutting the distance in half will reduce the
velocity by 2 , not 2. Therefore the speed will be greater than 6.0 m/s after traveling half the distance.
3. (c) Calculate the speed after
half the displacement:
๏ฆ v2 ๏ถ
v 2 12.0 m/s
v ๏ฝ v02 ๏ซ 2a ๏จ 12 ๏ x ๏ฉ ๏ฝ v02 ๏ซ a ๏ง ๏ญ 0 ๏ท ๏ฝ 0 ๏ฝ
๏ฝ 8.49 m/s
2
2
๏จ 2a ๏ธ
Insight: For constant acceleration, the velocity changes linearly with time, but nonlinearly with distance.
56. Picture the Problem: You drive in a straight line and then slow down to a stop.
Strategy: Use the constant acceleration equation of motion (Equation 2-7) to find the time. Once the time is known, we
can use the same equation to find the speed. In this case, the acceleration is negative because the car is slowing down.
t๏ฝ
Solution: 1. (a) Calculate the stopping time:
v ๏ญ v0 0 ๏ญ 16 m/s
๏ฝ
๏ฝ 5.0 s
a
๏ญ3.2 m/s 2
2. (b) Because the velocity varies linearly with time for constant acceleration, the velocity will be half the initial
velocity when you have braked for half the time. Therefore the speed after braking 2.5 s will be equal to 8.0 m/s.
v ๏ฝ v0 ๏ซ at ๏ฝ 16 m/s ๏ซ ๏จ ๏ญ3.2 m/s2 ๏ฉ ๏จ 2.5 s ๏ฉ ๏ฝ 8.0 m/s
3. (c) Calculate the speed after half the time:
Insight: For constant acceleration, the velocity changes linearly with time, but nonlinearly with distance.
Copyright ยฉ 2017 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No
portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
2 โ 21
James S. Walker, Physics, 5th Edition
Chapter 2: One-Dimensional Kinematics
57. Picture the Problem: A chameleonโs tongue accelerates in a straight line until it is extended to its full length.
Strategy: Employ the relationship between acceleration, displacement, and time (Equation 2-11) to find the
acceleration. Let the initial velocity v0 and the initial position x0 of the tongue each be zero.
Solution: 1. (a) Let x0 ๏ฝ v0 ๏ฝ 0 and calculate the acceleration:
a๏ฝ
2 x 2 ๏จ 0.16 m ๏ฉ
๏ฝ
๏ฝ 32 m/s 2
2
t2
0.10
s
๏จ
๏ฉ
2. (b) Because the displacement varies with the square of the time for constant acceleration, the displacement will be
less than half its final value when half the time has elapsed. Most of the displacement occurs when the tongue's speed is
greatest, late in the time interval. Therefore we expect the tongue to have extended less than 8.0 cm after 0.050 s.
x ๏ฝ 12 a t 2 ๏ฝ 12 ๏จ 32 m/s2 ๏ฉ ๏จ 0.050 s ๏ฉ ๏ฝ 4.0 cm
2
3. (c) Calculate the position of the tongue after half the time:
Insight: For constant acceleration, the displacement changes nonlinearly with both time and velocity. Notice that the
acceleration of the chameleonโs tongue is over three times the acceleration of gravity!
58. Picture the Problem: David Purley travels in a straight line, slowing down at a uniform rate until coming to rest.
Strategy: Use the time-free relationship between displacement, velocity, and acceleration (Equation 2-12) to find the
acceleration.
Solution: Calculate the acceleration:
0.278 m/s ๏ถ
๏ฆ
02 ๏ญ ๏ง173 km/h ๏ด
2
2
๏ท
v ๏ญ v0
1 km/h ๏ธ
๏จ
a๏ฝ
๏ฝ
2๏x
2 ๏จ 0.66 m ๏ฉ
a ๏ฝ ๏ญ1800 m/s 2 ๏ด
1.00 g
๏ฝ ๏ญ180 g
9.81 m/s 2
2
๏ a ๏ฝ 180 g
Insight: Mr. Purley was lucky to escape death when experiencing an acceleration this large! Weโll learn in Chapter 5
that a large acceleration implies a large force, which in this case must have been applied to his body in just the right way
to produce a non-lethal injury.
59. Picture the Problem: A boat slows down at a uniform rate as it coasts in a straight line.
Strategy: Because the initial and final velocities are known, the time can be determined from the average velocity and
the distance traveled. Then use the constant acceleration equation of motion (Equation 2-7) to find the acceleration and
the time-free equation (Equation 2-12) to find the velocity after the boat had coasted half the distance.
๏x
12 m
๏ฝ1
๏ฝ 5.7 s
2 ๏จ v ๏ซ v0 ๏ฉ
2 ๏จ1.6 ๏ซ 2.6 m/s ๏ฉ
Solution: 1. (a) Use the displacement and the
average velocity to find the time elapsed:
t๏ฝ 1
2. (b) Apply the definition of acceleration:
v ๏ญ v0 1.6 ๏ญ 2.6 m/s
๏ฝ
๏ฝ ๏ญ 0.175 m/s2 where the negative
t
5.7 s
sign means opposite the direction of motion.
3. (c) Calculate the velocity after coasting 6.0 m
using the time-free equation of motion:
a๏ฝ
v 2 ๏ฝ v02 ๏ซ 2 a ๏ x
v๏ฝ
๏จ 2.6 m/s ๏ฉ ๏ซ 2 ๏จ ๏ญ 0.175 m/s2 ๏ฉ ๏จ 6.0 m ๏ฉ ๏ฝ 2.2 m/s
2
Insight: For constant acceleration, the velocity changes linearly with time but nonlinearly with distance. That is why
the 2.2-m/s velocity after coasting 6.0 m is greater than the 2.1-m/s average speed the boat has over the entire 12-m
distance it coasted.
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portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
2 โ 22
James S. Walker, Physics, 5th Edition
Chapter 2: One-Dimensional Kinematics
60. Picture the Problem: A model rocket accelerates straight upward at a constant rate.
Strategy: Because the initial and final velocities are known, the time can be determined from the average velocity and
the distance traveled. The constant acceleration equation of motion (Equation 2-7) can then be used to find the
acceleration. Once that is known, the position of the rocket as a function of time is given by Equation 2-11, and the
velocity as a function of time is given by Equation 2-7.
๏x
4.2 m
๏ฝ1
๏ฝ 0.323 s ๏ฝ 0.32 s
2 ๏จ v ๏ซ v0 ๏ฉ
2 ๏จ 0 ๏ซ 26.0 m/s ๏ฉ
Solution: 1. (a) Use the displacement and the
average velocity to find the time elapsed:
t๏ฝ 1
2. (b) Apply the definition of acceleration:
a๏ฝ
3. (c) Find the rocketโs height, assuming x0 ๏ฝ v0 ๏ฝ 0 :
x ๏ฝ 12 a t 2 ๏ฝ 12 ๏จ80 m/s2 ๏ฉ ๏จ 0.10 s ๏ฉ ๏ฝ 0.40 m
4. Find the velocity of the rocket, assuming v0 ๏ฝ 0 :
v ๏ฝ 0 ๏ซ a t ๏ฝ ๏จ80 m/s2 ๏ฉ ๏จ 0.10 s ๏ฉ ๏ฝ 8.0 m/s
v ๏ญ v0 26.0 ๏ญ 0 m/s
๏ฝ
๏ฝ 80 m/s 2
t
0.323 s
2
Insight: Model rockets accelerate at very large rates, but only for a very short time. Still, even inexpensive starter
rockets can reach 1500 ft in altitude and can be great fun to build and launch!
61. Picture the Problem: The infamous chicken dashes toward home plate while playing baseball, and then slides along a
straight line and comes to rest.
Strategy: Because the initial and final velocities and the time elapsed are known, the acceleration can be determined
from the constant acceleration equation of motion (Equation 2-7). The distance traveled can be found from the average
velocity and the time elapsed (Equation 2-10).
Solution: 1. (a) Calculate the acceleration:
v ๏ญ v0 0 ๏ญ 5.7 m/s
๏ฝ
๏ฝ ๏ญ 4.8 m/s2 , where the negative sign
t
1.2 s
means opposite the direction of motion, or toward third base.
2. (b) Use the average velocity and time
to find the distance the chicken slides:
๏ x ๏ฝ 12 ๏จ v ๏ซ v0 ๏ฉ t ๏ฝ 12 ๏จ 0 ๏ซ 5.7 m/s ๏ฉ๏จ1.2 s ๏ฉ ๏ฝ 3.4 m
a๏ฝ
Insight: If the dirt had accelerated the chicken at a lesser rate, the chicken would have had nonzero speed as it crossed
home plate. A larger magnitude acceleration would stop the chicken before reaching the plate, and it would be out!
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2 โ 23
James S. Walker, Physics, 5th Edition
Chapter 2: One-Dimensional Kinematics
62. Picture the Problem: The distance-versus-time plot at right
shows how a bicyclist can overtake his friend by pedaling at constant
acceleration.
Strategy: To find the time elapsed when the two bicyclists meet, we must
set the constant velocity equation of motion of the friend (Equation 2-8)
equal to the constant acceleration equation of motion (Equation 2-11) of
the bicyclist. Once the time is known, the displacement and velocity of
the bicyclist can be determined from Eqs. 2-10 and 2-7, respectively.
xfriend ๏ฝ xbicyclist
Solution: 1. (a) Set the two equations of
motion equal to each other. For the friend,
use Equation 2-8 with x0 ๏ฝ 0 and for the
bicyclist, use Equation 2-11 with x0 ๏ฝ 0 and v0 ๏ฝ 0 :
vfriend t ๏ฝ 0 ๏ซ 0 ๏ซ 12 abicyclist ๏จ t ๏ญ 2 ๏ฉ
2
vfriend t ๏ฝ 12 abicyclist ๏จ t 2 ๏ญ 4t ๏ซ 4 ๏ฉ
2. Solve the two equations for t by rearranging
them into a quadratic expression:
๏ฉ
๏น
2 ๏จ 3.5 m/s ๏ฉ ๏น
๏ฉ
2v
0 ๏ฝ t 2 ๏ญ ๏ช 4 ๏ซ friend ๏บ t ๏ซ 4 ๏ฝ t 2 ๏ญ ๏ช 4 ๏ซ
๏บt ๏ซ 4
abicyclist ๏บ๏ป
2.4 m/s 2 ๏ป
๏ช๏ซ
๏ซ
0 ๏ฝ t 2 ๏ญ 6.92t ๏ซ 4
t๏ฝ
3. Now use the quadratic formula:
๏ซ 6.92 ๏ฑ
6.922 ๏ญ 4 ๏จ1๏ฉ๏จ 4 ๏ฉ
2
๏ฝ 6.3, 0.64 s
4. We choose the larger root because the time must be greater than 2.0 s, the time at which the bicyclist began pursuing
his friend. The bicyclist will overtake his friend 6.3 s after his friend passes him.
xfriend ๏ฝ v0 t ๏ฝ ๏จ 3.5 m/s ๏ฉ๏จ 6.3 s ๏ฉ ๏ฝ 22 m
5. (b) Use the known time to find the position:
xbicyclist ๏ฝ 12 abicyclist ๏จ t ๏ญ 2 ๏ฉ ๏ฝ 12 ๏จ 2.4 m/s 2 ๏ฉ ๏จ 4.3 s ๏ฉ ๏ฝ 22 m
2
6. (c) Use Equation 2-7 to find vbicyclist. Keep in mind
that v0 ๏ฝ 0 and that the bicyclist doesnโt begin
accelerating until two seconds have elapsed:
2
v ๏ฝ 0 ๏ซ abicyclist ๏จ t ๏ญ 2 ๏ฉ ๏ฝ ๏จ 2.4 m/s2 ๏ฉ ๏จ 6.3 ๏ญ 2.0 s ๏ฉ ๏ฝ 10 m/s
Insight: Even a smaller acceleration would allow the bicyclist to catch up to the friend, because the speed is always
increasing for any nonzero acceleration. Hence the bicyclistโs speed would eventually exceed the friendโs speed and the
two would meet some time after that.
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portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
2 โ 24
James S. Walker, Physics, 5th Edition
Chapter 2: One-Dimensional Kinematics
63. Picture the Problem: The velocity-versus-time plot at right indicates the
velocity of a car as it accelerates in the forward direction, maintains a
constant speed, and then rapidly slows down to a stop.
Strategy: The distance traveled by the car is equal to the area under the
velocity-versus-time plot. Because the distance traveled is known to be
22 m, we can use that fact to determine the unknown speed V. Once we
know the velocity as a function of time we can answer any other question
about its motion during the time interval.
Solution: 1. (a) Determine the area under the curve
by adding the area of the triangle from 0 to 4 s, the
rectangle from 4 to 6 s, and the triangle from 6 to 8 s.
x ๏ฝ 12 ๏จ 4 ๏ญ 0 s ๏ฉV ๏ซ ๏จ 6 ๏ญ 4 s ๏ฉV ๏ซ 12 ๏จ8 ๏ญ 6 s ๏ฉV ๏ฝ ๏จ5 s ๏ฉV
2. Set x equal to 22 m and solve for V:
x ๏ฝ ๏จ 5.0 s ๏ฉV ๏ฝ 22 m ๏ V ๏ฝ ๏จ 22 / 5.0 ๏ฉ m/s ๏ฝ 4.4 m/s
3. Now find the area of the triangle from 0 to 4 s:
x1 ๏ฝ 12 ๏จ 4 ๏ญ 0 s ๏ฉ๏จ 4.4 m/s ๏ฉ ๏ฝ 8.8 m
4. (b) Find the area of the triangle from 6 to 8 s:
x1 ๏ฝ 12 ๏จ8 ๏ญ 6 s ๏ฉ๏จ 4.4 m/s ๏ฉ ๏ฝ 4.4 m
5. (c) We found the unknown speed in step 2:
V ๏ฝ 4.4 m/s
Insight: The velocity-versus-time graph is a rich source of information. Besides velocity and time information, you can
determine acceleration from the slope of the graph and distance traveled from the area under the graph.
64. Picture the Problem: The velocity-versus-time plots of the car and the
truck are shown at right. The car begins with a positive position and a
negative velocity, so it must be represented by the lower line. The truck
begins with a negative position and a positive velocity, so it is represented
by the upper line.
Strategy: The distances traveled by the car and the truck are equal to the
areas under their velocity-versus-time plots. We can determine the
distances traveled from the plots and use the known initial positions to
find the final positions and the final separation.
Solution: 1. Find the final position of
the truck. The truckโs displacement
๏ xtruck is the area under its v vs. t graph:
xtruck ๏ฝ x0,truck ๏ซ ๏ xtruck ๏ฝ ๏จ ๏ญ35 m ๏ฉ ๏ซ 12 ๏จ 2.5 ๏ญ 0 s ๏ฉ๏จ10 m/s ๏ฉ ๏ฝ ๏ญ22.5 m
2. Find the final position of the car. The
carโs displacement ๏ xcar is the area
under its v vs. t graph:
xcar ๏ฝ x0,car ๏ซ ๏ xcar ๏ฝ ๏จ15 m๏ฉ ๏ซ 12 ๏จ 3.5 ๏ญ 0 s ๏ฉ๏จ ๏ญ15 m/s ๏ฉ ๏ฝ ๏ญ11.25 m
3. Now find the separation:
xcar ๏ญ xtruck ๏ฝ ๏จ ๏ญ11.25 m ๏ฉ ๏ญ ๏จ ๏ญ22.5 m ๏ฉ ๏ฝ 11.3 m
Insight: The velocity-versus-time graph is a rich source of information. Besides velocity and time information, you can
determine acceleration from the slope of the graph, and distance traveled from the area under the graph. In this case, we
can see the acceleration of the car (4.29 m/s2) has a greater magnitude than the acceleration of the truck (โ4.00 m/s 2).
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portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
2 โ 25
James S. Walker, Physics, 5th Edition
Chapter 2: One-Dimensional Kinematics
65. Picture the Problem: Penguins slide down three different frictionless ramps, A, B, and C. The distance along each
ramp and the average sliding times are recorded.
Strategy: Use the relationship between distance, acceleration, and time (Equation 2-11) to determine the accelerations
of the penguins. Then use a ๏ฝ g sin ๏ฑ to determine the angle of inclination ๏ฑ for each ramp.
Solution: 1. (a) Write an expression for the
acceleration, assuming that x0 ๏ฝ v0 ๏ฝ 0 :
x ๏ฝ 0 ๏ซ 0 ๏ซ 12 at 2
2. Calculate the acceleration of penguins
that slide along ramp A:
aA ๏ฝ
2 xA 2 ๏จ 4.09 m ๏ฉ
๏ฝ
๏ฝ 1.71 m/s 2
2
tA2
๏จ 2.19 s ๏ฉ
3. Calculate the acceleration of penguins
that slide along ramp B:
aB ๏ฝ
2 xB 2 ๏จ1.96 m ๏ฉ
๏ฝ
๏ฝ 3.36 m/s 2
2
2
tB
๏จ1.08 s ๏ฉ
4. Calculate the acceleration of penguins
that slide along ramp C:
aC ๏ฝ
2 xC 2 ๏จ1.08 m ๏ฉ
๏ฝ
๏ฝ 4.91 m/s 2
2
tC2
๏จ 0.663 s ๏ฉ
5. (b) Write an expression for the angle of
incline ๏ฑ :
a ๏ฝ g sin ๏ฑ
6. Calculate the angle of incline for ramp A:
๏ฑ A ๏ฝ sin ๏ญ1 ๏ง
7. Calculate the angle of incline for ramp B:
๏ฑ B ๏ฝ sin ๏ญ1 ๏ง
6. Calculate the angle of incline for ramp C:
๏ฑC ๏ฝ sin ๏ญ1 ๏ง
๏ a ๏ฝ 2x t 2
๏ฆa๏ถ
๏ ๏ฑ ๏ฝ sin ๏ญ1 ๏ง ๏ท
๏จg๏ธ
๏ฆ 1.71 m/s 2 ๏ถ
๏ฝ 10.0๏ฐ
2 ๏ท
๏จ 9.81 m/s ๏ธ
๏ฆ 3.36 m/s 2 ๏ถ
๏ฝ 20.0๏ฐ
2 ๏ท
๏จ 9.81 m/s ๏ธ
๏ฆ 4.91 m/s 2 ๏ถ
๏ฝ 30.0๏ฐ
2 ๏ท
๏จ 9.81 m/s ๏ธ
Insight: Along a steeper ramp there is a greater component of gravitational force that is parallel to the ramp, resulting in
a larger acceleration.
66. Picture the Problem: Two balls are each thrown with speed v0 from the same initial height. Ball 1 is thrown straight
upward and ball 2 is thrown straight downward.
Strategy: Use the known set of kinematic equations that describe motion with constant acceleration to determine the
relative speeds of balls 1 and 2 when they hit the ground.
Solution: 1. Solve Equation 2-12 for v1,
assuming the ball is thrown upward with
velocity v0 :
v ๏ฝ v02 ๏ซ 2 ๏จ ๏ญ g ๏ฉ ๏ x ๏ฝ v02 ๏ญ 2 g ๏ x
2. Solve Equation 2-12 for v2, assuming the ball
is thrown downward with velocity v0 :
v๏ฝ
๏จ ๏ญv0 ๏ฉ ๏ซ 2 ๏จ ๏ญ g ๏ฉ ๏ x ๏ฝ v02 ๏ญ 2 g ๏ x
2
3. By comparing the two expressions for v above we can conclude that the best answer is B. The speed of ball 1 is equal
to the speed of ball 2.
Insight: In a later chapter weโll come to the same conclusion from an understanding of the conservation of mechanical
energy. The balls have the same speed just before they land because they both have the same downward speed when
they are at the level of the roof. Ball 2 simply starts off with the speed v0 downward. Ball 1 travels upward initially, but
when it returns to the level of the roof it is moving downward with the speed v0 , just like ball 2.
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portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
2 โ 26
James S. Walker, Physics, 5th Edition
Chapter 2: One-Dimensional Kinematics
67. Picture the Problem: A cliff diver drops from rest, picking up speed with the acceleration of gravity.
Strategy: Convert the speed in mi/h to m/s, and then solve Equation 2-7 for the time required for the cliff diver to reach
that speed when she accelerates at 9.81 m/s2.
mi
1 m/s
๏ด
๏ฝ 26.8 m/s
h 2.24 mi/h
Solution: 1. Convert v to units of m/s2:
60
2. Solve Equation 2-7 for the time:
v ๏ฝ v0 ๏ซ g t ๏ฝ 0 ๏ซ g t
t๏ฝ
v 26.8 m/s
๏ฝ
๏ฝ 2.7 s
g 9.81 m/s 2
Insight: This is significantly less than the 3.5 s required for a powerful car to achieve 60.0 mi/h from rest.
68. Picture the Problem: A juggler throws a ball straight upward and later catches it at the same height it was thrown.
Strategy: Use the known acceleration of gravity (9.81 m/s2) and Equation 2-7 to find the initial speed of the ball,
assuming by symmetry that the final speed is the same as the initial speed, except the final velocity is downward
(negative).
๏ญv0 ๏ฝ v0 ๏ญ g t ๏ g t ๏ฝ 2v0
Solution: Solve Equation 2-7
for the initial velocity:
v0 ๏ฝ 12 g t ๏ฝ 12 ๏จ 9.81 m/s 2 ๏ฉ ๏จ 3.2 s ๏ฉ ๏ฝ 16 m/s
Insight: This speed is equivalent to 35 mi/h, a reasonably easy throw for an accomplished juggler.
69. Picture the Problem: Snowboarder Shaun White soars straight upward a distance 6.4 m above the rim of a half-pipe.
Strategy: Because the height of the snowboard and rider is known, the time-free equation of motion (velocity in terms
of displacement, Equation 2-12) can be used to find the takeoff speed.
Solution: Solve the time-free
equation of motion for v0 :
v0 ๏ฝ v 2 ๏ญ 2 g ๏ x ๏ฝ 02 ๏ญ 2 ๏จ ๏ญ9.81 m/s2 ๏ฉ ๏จ 6.4 m ๏ฉ ๏ฝ 11 m/s
Insight: That speed is about 25 mi/h straight upward! Olympic snowboarders must be very athletic as well as acrobatic
to perform the feats we witness during the Games.
70. Picture the Problem: A gull drops a clam shell, which falls from rest straight down under the influence of gravity.
Strategy: Because the distance of the fall is known, use the time-free equation of motion (velocity in terms of
displacement, Equation 2-12) to find the landing speed.
Solution: Solve the time-free equation of
motion for v. Let v0 ๏ฝ 0 and let downward
be the positive direction.
v ๏ฝ v02 ๏ซ 2 g ๏ x ๏ฝ 02 ๏ซ 2 ๏จ 9.81 m/s2 ๏ฉ ๏จ17 m ๏ฉ ๏ฝ 18 m/s
Insight: That speed (about 41 mi/h) is sufficient to shatter the shell and provide a tasty meal!
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portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
2 โ 27
James S. Walker, Physics, 5th Edition
Chapter 2: One-Dimensional Kinematics
71. Picture the Problem: A volcano launches a lava bomb straight upward. It slows down under the influence of gravity,
coming to rest momentarily before falling downward.
Strategy: Because the acceleration of gravity is known, the constant acceleration equation of motion (velocity as a
function of time, Equation 2-7) can be used to find the speed and velocity as a function of time. Let upward be the
positive direction.
Solution: 1. (a) Apply Equation 2-7
directly with a = โg:
v ๏ฝ v0 ๏ญ g t ๏ฝ 28 m/s ๏ญ ๏จ 9.81 m/s2 ๏ฉ ๏จ 2.0 s ๏ฉ ๏ฝ 8.4 m/s
2. (b) Apply Equation 2-7 directly with a = โg:
v ๏ฝ v0 ๏ญ g t ๏ฝ 28 m/s ๏ญ ๏จ 9.81 m/s2 ๏ฉ ๏จ 3.0 s ๏ฉ ๏ฝ ๏ญ1.4 m/s
3. The positive sign for the velocity in part (a) indicates that the lava bomb is traveling upward, and the negative sign
for part (b) means it is traveling downward.
Insight: We can see the lava bomb must have reached its peak between 2.0 and 3.0 seconds. In fact, it reached it at
t ๏ฝ ๏จ v ๏ญ v0 ๏ฉ a ๏ฝ ๏จ 0 ๏ญ 28 m/s ๏ฉ ๏จ ๏ญ9.81 m/s2 ๏ฉ ๏ฝ 2.85 s.
72. Picture the Problem: Volcanic material on Io travels straight upward, slowing down under the influence of gravity
until it momentarily comes to rest at its maximum altitude.
Strategy: Because the maximum altitude is known, use the time-free equation of motion (velocity in terms of
displacement, Equation 2-12) to find the initial velocity. Let upward be the positive direction, so that a = โ1.80 m/s2.
Solution: Solve the time-free
equation of motion for v0 :
v0 ๏ฝ v 2 ๏ญ 2 a ๏ x ๏ฝ 02 ๏ญ 2 ๏จ ๏ญ1.80 m/s 2 ๏ฉ๏จ 3.00 ๏ด105 m ๏ฉ
๏ฝ 1040 m/s ๏ฝ 1.04 km/s
Insight: On Earth that speed would only hurl the material to an altitude of 55 km, as opposed to 300 km on Io. Still,
thatโs a very impressive initial velocity! It is equivalent to the muzzle velocity of a bullet, and is 2.5 times the speed of
sound on Earth.
73. Picture the Problem: A ruler falls straight down under the influence of gravity.
Strategy: Because the acceleration and initial velocity (zero) of the ruler are known, use the position as a function of
time equation of motion (Equation 2-11) to find the time.
Solution: Solve Equation 2-11 for t. Let v0 ๏ฝ 0
and let downward be the positive direction.
t๏ฝ
2 ๏จ 0.052 m ๏ฉ
2๏ x
๏ฝ
๏ฝ 0.10 s
g
9.81 m/s 2
Insight: This is a very good reaction time, about half the average human reaction time of 0.20 s.
74. Picture the Problem: A hammer drops straight downward and passes by two windows of equal height.
Strategy: Use the definition of acceleration together with the knowledge that a falling hammer undergoes constant
acceleration to answer the conceptual question.
Solution: 1. (a) The acceleration of the hammer is a constant throughout its flight (neglecting air friction) so its speed
increases by the same amount for each equivalent time interval. However, it passes by the second window in a smaller
amount of time than it took to pass by the first window because its speed has increased. We conclude that increase in
speed of the hammer as it drops past window 1 is greater than the increase in speed as it drops past window 2.
2. (b) The best explanation (see the discussion above) is III. The hammer spends more time dropping past window 1.
Statement 1 is false because acceleration is independent of speed, and statement II is false because acceleration is rate of
change of speed per time not distance.
Insight: If the hammer were thrown upward, its speed decrease as it passes window 2 would be less than the decrease in
its speed as it passes window 1, again because it is traveling slower as it passes window 1.
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portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
2 โ 28
James S. Walker, Physics, 5th Edition
Chapter 2: One-Dimensional Kinematics
75. Picture the Problem: A hammer drops straight downward and passes by two windows of equal height.
Strategy: The velocity-versus-time graph contains two pieces of information: the slope of the graph is the acceleration,
and the area under the graph is the distance traveled. Use this knowledge to answer the conceptual question.
Solution: 1. (a) The two windows have the same height, so the hammer travels the same distance as it passes each
window. We conclude that the area of the shaded region corresponding to window 1 is equal to the area of the shaded
region corresponding to window 2.
2. (b) The best explanation (see the discussion above) is II. The windows are equally tall. Statement I is true, but not
relevant, and statement III is true, but not relevant.
Insight: If the hammer were thrown upward, the velocity-versus-time graph would have a negative slope, but the
shaded areas corresponding to each window would still be equal, with the tall and narrow shaded area for window 2 on
the left (because the hammer passes it first) and the short and wide shaded area for window 1 on the right.
76. Picture the Problem: Two balls are thrown upward with the same initial speed but at different times. The second ball
is thrown at the instant the first ball has reached the peak of its flight.
Strategy: The average speed of the ball is smaller at altitudes above 12 h , so that it spends a greater fraction of time in
that region than it does at altitudes below 12 h . Use this insight to answer the conceptual question.
Solution: The second ball will reach 12 h on its way up sooner than the first ball will reach 12 h on its way down because
the speed of each ball is greater at low altitudes than at high altitudes. We conclude that the two balls pass at an altitude
that is above 12 h .
Insight: A careful analysis reveals that the two balls will pass each other at altitude of 34 h .
77. Picture the Problem: Several swimmers fall straight down from a bridge into the Snohomish River.
Strategy: The initial velocities of the swimmers are zero because they step off the bridge rather than jump up or dive
downward. Use the equation of motion for position as a function of time and acceleration, realizing that the acceleration
in each case is 9.81 m/s2. Set x0 ๏ฝ 0 and let downward be the positive direction for simplicity. The known
acceleration can be used to find velocity as a function of time for part (b). Finally, the same equation of motion for part
(a) can be solved for time in order to answer part (c).
Solution: 1. (a) Calculate the fall distance:
x ๏ฝ x0 ๏ซ v0 t ๏ซ 12 a t 2 ๏ฝ 0.0 m ๏ซ 0 ๏ซ 12 ๏จ 9.81 m/s 2 ๏ฉ ๏จ1.5 s ๏ฉ
2
x ๏ฝ 11 m
2. (b) Calculate the final speed if v0 ๏ฝ 0:
v ๏ฝ v0 ๏ซ a t ๏ฝ 0 ๏ซ ๏จ 9.81 m/s2 ๏ฉ ๏จ1.5 s ๏ฉ ๏ฝ 15 m/s
3. (c) Calculate the fall time for twice the distance:
t๏ฝ
2 ๏จ11 m ๏ด 2 ๏ฉ
2x
๏ฝ
๏ฝ 2.1 s
a
9.81 m/s 2
Insight: The time in part (c) doesnโt double because it depends upon the square root of the distance the swimmer falls.
If you want to double the fall time you must quadruple the height of the bridge!
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portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
2 โ 29
James S. Walker, Physics, 5th Edition
Chapter 2: One-Dimensional Kinematics
78. Picture the Problem: Water in the highest fountain is projected with a large upward velocity, rises straight upward, and
momentarily comes to rest before falling straight back down again.
Strategy: By analyzing the time-free equation of motion (Equation 2-12) with v ๏ฝ 0 (because the water briefly comes to
rest at the top of its trajectory), we can see that the initial velocity v0 increases with the square root of the fountain
height. The known fountain height and acceleration of gravity can also be used to determine the time it takes for the
water to reach the peak using the position as a function of time (Equation 2-11).
Solution: 1. (a) Calculate v0 assuming the
water comes to rest ( v ๏ฝ 0 ) at the top:
02 ๏ฝ v02 ๏ญ 2 g ๏x
v0 ๏ฝ
2 g ๏x ๏ฝ
2. (b) Calculate the time required for the
water to reach the top of the fountain:
t๏ฝ
2x
๏ฝ
a
2 ๏จ 9.81 m/s 2 ๏ฉ ๏จ 560 ft ๏ด 0.305 m/ft ๏ฉ ๏ฝ 58 m/s
2 ๏จ 560 ft ๏ด 0.305 m/ft ๏ฉ
9.81 m/s 2
๏ฝ 5.9 s
Insight: The speed of 58 m/s corresponds to 130 mi/h. The fountain is produced by a world-class water pump!
79. Picture the Problem: A basketball bounces straight up, momentarily comes to rest, and then falls straight back down.
Strategy: If air friction is neglected, the time it takes the ball to fall is the same as the time it takes the ball to rise.
Therefore, the maximum height of the ball is also the distance a ball will fall for 1.6 s. Use the equation of motion for
position as a function of time and acceleration, realizing that the acceleration in each case is 9.81 m/s2. Set x0 ๏ฝ v0 ๏ฝ 0
and let downward be the positive direction for simplicity.
Solution: Calculate the maximum height:
x ๏ฝ x0 ๏ซ v0 t ๏ซ 12 a t 2 ๏ฝ 0.0 m ๏ซ 0 ๏ซ 12 ๏จ 9.81 m/s2 ๏ฉ ๏จ1.6 s ๏ฉ ๏ฝ 12.6 m ๏ฝ 13 m
2
Insight: The 12.6-m height corresponds to 41 ft. The ball must have rebounded from the floor with a speed of 15.7 m/s
or 35 mi/h. The player was pretty angry!
80. Picture the Problem: A baseball glove rises straight up, momentarily comes to rest, and then falls straight back down.
Strategy: The glove will land with the same speed it was released, neglecting any air friction, so the final velocity
v ๏ฝ ๏ญ 6.5 m/s. We can use the equation of motion for velocity as a function of time (Eq. 2-7) to find the time of flight.
Solution: 1. (a) Calculate the total time of flight
t๏ฝ
v ๏ญ v0 ๏จ ๏ญ 6.5๏ฉ ๏ญ ๏จ 6.5๏ฉ m/s
๏ฝ
๏ฝ 1.3 s
a
๏ญ9.81 m/s 2
2. (b) Calculate the time to reach maximum height:
t๏ฝ
v ๏ญ v0 0 ๏ญ 6.5 m/s
๏ฝ
๏ฝ 0.66 s
a
๏ญ9.81 m/s 2
Insight: Throwing the glove upward with twice the speed will double the time of flight, but the maximum height
attained by the glove (2.15 m for a 6.5 m/s initial speed) will increase by only a factor of 2 .
Copyright ยฉ 2017 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No
portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
2 โ 30
James S. Walker, Physics, 5th Edition
Chapter 2: One-Dimensional Kinematics
81. Picture the Problem: Two balls fall straight down under the influence of gravity. The first ball falls from rest but the
second ball is given an initial downward velocity.
Strategy: Because the fall distance is known in each case, use the velocity in terms of displacement equation of motion
(Equation 2-12) to predict the final velocity. Let downward be the positive direction for simplicity.
Solution: 1. (a) The speed increases linearly with time but nonlinearly with distance. Because the first ball has a lower
initial velocity and hence a lower average velocity, it spends more time in the air. The first (dropped) ball will
therefore experience a larger increase in speed.
2. (b) First ball: Solve Eq. 2-12
for v, setting v0 ๏ฝ 0 :
v ๏ฝ 02 ๏ซ 2 g ๏ x ๏ฝ 2 ๏จ 9.81 m/s2 ๏ฉ ๏จ 30.5 m ๏ฉ ๏ฝ 24.5 m/s
๏จ11.2 m/s ๏ฉ ๏ซ 2 ๏จ9.81 m/s2 ๏ฉ ๏จ30.5 m ๏ฉ ๏ฝ 26.9 m/s
3. Second ball: Solve Eq. 2-12 for v:
v ๏ฝ v02 ๏ซ 2 g ๏ x ๏ฝ
4. Compare the ๏ v values:
๏v1 ๏ฝ 24.5 ๏ญ 0 m/s ๏ฝ 24.5 m/s for the first ball and
2
๏v2 ๏ฝ 26.9 ๏ญ 11.2 m/s ๏ฝ 15.7 m/s for the second ball.
Insight: The second ball is certainly going faster, but its change in speed is less than the first ball.
82. Picture the Problem: An arrow rises straight upward, slowing down due to the acceleration of gravity.
Strategy: Because the position, time, and acceleration are all known, we can use the equation of motion for position as
a function of time (Equation 2-11) to find the initial velocity v0 . The same equation could be used to find the time
required to rise to a height of 15.0 m above its launch point. Let the launch position be x0 ๏ฝ 0 and let upward be the
positive direction.
x ๏ญ 12 a t 2 30.0 m ๏ญ 2 ๏จ ๏ญ9.81 m/s ๏ฉ ๏จ 2.00 s ๏ฉ
๏ฝ
๏ฝ 24.8 m/s
t
2.00 s
2
1
Solution: 1. (a) Calculate v0 from a
rearrangement of Equation 2-11:
v0 ๏ฝ
2. (b) Solve Equation 2-11 with x = 15.0 m:
15.0 m ๏ฝ ๏จ 24.8 m/s ๏ฉ t ๏ญ 12 ๏จ 9.81 m/s 2 ๏ฉ t 2
2
0 ๏ฝ ๏จ ๏ญ 4.905 m/s 2 ๏ฉ t 2 ๏ซ ๏จ 24.8 m/s ๏ฉ t ๏ญ 15.0 m
3. Now use the quadratic formula:
๏ญb ๏ฑ b 2 ๏ญ 4ac ๏ญ24.8 ๏ฑ
๏ฝ
2a
t ๏ฝ 0.702 s , 4.36 s
t๏ฝ
๏จ 24.8๏ฉ ๏ญ 4 ๏จ ๏ญ 4.905๏ฉ๏จ ๏ญ15.0 ๏ฉ
2
๏ญ9.81
Insight: The second root of the solution to part (b) corresponds to the time when the arrow, after rising to its maximum
height, falls back to a position 15.0 m above the launch point.
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portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
2 โ 31
James S. Walker, Physics, 5th Edition
Chapter 2: One-Dimensional Kinematics
83. Picture the Problem: A book accelerates straight downward and hits the floor of an elevator that is descending at
constant speed.
Strategy: The constant speed motion of the elevator does not affect the acceleration of the book. From the perspective
of an observer outside the elevator, both the book and the floor have an initial downward velocity of 3.0 m/s. Therefore,
from your perspective the motion of the book is no different than if the elevator were at rest. Solve the position as a
function of time and acceleration equation (Equation 2-11) for t, setting v0 ๏ฝ 0 and letting downward be the positive
direction. Then use velocity as a function of time (Equation 2-7) to find the speed of the book when it lands.
2 ๏จ1.2 m ๏ฉ
2x
๏ฝ
๏ฝ 0.49 s
g
9.81 m/s 2
Solution: 1. (a) Solve Equation 2-11
for t, setting x0 ๏ฝ v0 ๏ฝ 0 :
t๏ฝ
2. (b) Apply Equation 2-7 to find v:
v ๏ฝ v0 ๏ซ g t ๏ฝ 0 ๏ซ ๏จ 9.81 m/s2 ๏ฉ ๏จ 0.49 s ๏ฉ ๏ฝ 4.8 m/s
Insight: The speed in part (b) is relative to you. Relative to the ground the velocity of the book is 4.8 + 3.0 = 7.8 m/s in
the downward direction.
84. Picture the Problem: A camera has an initial downward velocity of 2.3 m/s when it is dropped from a hot-air balloon.
The camera accelerates straight downward before striking the ground.
Strategy: One way to solve this problem is to use the quadratic formula to find t from the position as a function of time
and acceleration equation (Equation 2-11). Then the definition of acceleration can be used to find the final velocity.
Hereโs another way: Find the final velocity from the time-free equation of motion (Equation 2-12) and use the
relationship between average velocity, position, and time (Equation 2-10) to find the time. Weโll therefore be solving
this problem backwards, finding the answer to (b) first and then (a). Let upward be the positive direction, so that
v0 ๏ฝ ๏ญ2.3 m/s and ๏ x ๏ฝ x ๏ญ x0 ๏ฝ 0 ๏ญ 41 m ๏ฝ ๏ญ 41 m.
Solution: 1. (a) Solve Equation 2-12 for v:
v ๏ฝ v02 ๏ซ 2 g ๏ x ๏ฝ
2. Solve Equation 2-10 for t:
t๏ฝ 1
3. (b) We found v in step 1:
v ๏ฝ ๏ญ28 m/s
๏จ ๏ญ2.3 m/s ๏ฉ ๏ซ 2 ๏จ ๏ญ9.81 m/s2 ๏ฉ ๏จ ๏ญ 41 m ๏ฉ ๏ฝ ๏ญ28 m/s
2
๏x
๏ญ 41 m
๏ฝ 1
๏ฝ 2.7 s
2 ๏จ v ๏ซ v0 ๏ฉ
2 ๏จ ๏ญ28 ๏ญ 2.3 m/s ๏ฉ
Insight: There is often more than one way to approach constant acceleration problems, some easier than others. In this
case our strategy allowed us to avoid using the quadratic formula to find t.
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portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
2 โ 32
James S. Walker, Physics, 5th Edition
Chapter 2: One-Dimensional Kinematics
85. Picture the Problem: A model rocket rises straight upward, accelerating over a distance of 29 m and then slowing
down and coming to rest at some altitude higher than 29 m.
Strategy: Use the given acceleration and distance and the time-free equation of motion (Equation 2-12) to find the
velocity of the rocket at the end of its acceleration phase, when its altitude is 29 m. Use that as the initial velocity of the
free-fall stage in order to find the maximum altitude (Equation 2-12 again). Then apply Equation 2-12 a third time to
find the velocity of the rocket when it returns to the ground. The given and calculated positions at various stages of the
flight can then be used to find the elapsed time in each stage and the total time of flight.
Solution: 1. (a) Find the velocity at the end
of the boost phase using Equation 2-12:
vboost ๏ฝ v02 ๏ซ 2 g ๏ x ๏ฝ 02 ๏ซ 2 ๏จ12 m/s2 ๏ฉ ๏จ 29 m ๏ฉ ๏ฝ 26.4 m/s
2. Find the height change during the
boost phase using Equation 2-12 and a
final speed of zero:
2
02 ๏ฝ vboost
๏ญ 2 g ๏ xboost ๏ ๏ xboost ๏ฝ
3. Now find the overall maximum height:
v2
hmax ๏ฝ h0 ๏ซ ๏ xboost ๏ฝ 29 m ๏ซ boost
2g
๏ฝ 29 m ๏ซ
4. (b) Apply Equation 2-12 once again
between the end of the boost phase and
the point where it hits the ground:
๏จ 26.4 m/s ๏ฉ
2
vboost
2g
2
2 ๏จ 9.81 m/s 2 ๏ฉ
๏ฝ 29 ๏ซ 36 m ๏ฝ 65 m
2
v 2 ๏ฝ vboost
๏ญ 2g ๏ x
๏จ 26.4 m/s ๏ฉ ๏ญ 2 ๏จ 9.81 m/s 2 ๏ฉ ๏จ ๏ญ29 m ๏ฉ
2
v ๏ฝ vboost
๏ญ 2g๏ x ๏ฝ
2
๏ฝ 36 m/s
๏ xboost
5. (c) First find the duration of the boost
phase. Use the known positions and
Equation 2-10:
tboost ๏ฝ 1
6. Now find the time for the rocket to
reach its maximum altitude from the end
of the boost phase:
tup ๏ฝ
7. Now find the time for the rocket to
fall back to the ground:
tdown ๏ฝ
8. Sum the times to find the time of flight:
ttotal ๏ฝ tboost ๏ซ tup ๏ซ tdown ๏ฝ 2.2 ๏ซ 2.7 ๏ซ 3.6 s ๏ฝ 8.5 s
2 ๏จ v0 ๏ซ vboost ๏ฉ
๏ xboost
1
2
๏จv
boost
๏ซ vtop ๏ฉ
๏จv ๏ซ v
top
2 ๏จ 0 ๏ซ 26.4 m/s ๏ฉ
๏ฝ 2.2 s
36 m
๏ฝ 2.7 s
2 ๏จ 26.4 ๏ซ 0 m/s ๏ฉ
๏ฝ1
๏ xdown
1
2
29 m
๏ฝ 1
ground
๏ฉ
65 m
๏ฝ 3.6 s
0
๏ซ
36 m/s ๏ฉ
2๏จ
๏ฝ1
Insight: Notice how knowledge of the initial and final velocities in each stage, and the distance traveled in each stage,
allowed the calculation of the elapsed times using the relatively simple Equation 2-10, as opposed to the quadratic
Equation 2-11. Learning to recognize the easiest route to the answer is an important skill to obtain.
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portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
2 โ 33
James S. Walker, Physics, 5th Edition
Chapter 2: One-Dimensional Kinematics
86. Picture the Problem: The vertical position-versus-time plot of a flying
squirrel is shown at right. The squirrel starts from rest and drops a
distance of x0 = 4.0 m in t = 1.10 s.
Strategy: Because the squirrel starts from rest and lands at x = 0, the
equation of motion for position as a function of time (Equation 2-11) can
be solved to find the acceleration a. We expect the acceleration to be
negative because the squirrel begins from a positive height and ends at a
zero height. The negative slope of the plot also indicates the velocity of
the squirrel is downward and increasing in magnitude.
Solution: 1. Solve the position as a function
of time equation for the acceleration a:
2. Calculate the squirrelโs acceleration:
x ๏ฝ x0 ๏ซ v0 t ๏ซ 12 a t 2
0 ๏ฝ x0 ๏ซ 0 ๏ซ 12 a t 2
a๏ฝ๏ญ
๏
๏ญ2 x0
๏ฝa
t2
2 ๏จ 4.0 m ๏ฉ
2 x0
๏ฝ๏ญ
๏ฝ ๏ญ 6.6 m/s 2
2
2
t
๏จ1.10 s ๏ฉ
Insight: If the squirrel did not have a patagium to slow its descent, its acceleration would be close to โ9.81 m/s2.
87. Picture the Problem: The height-versus-time plot of a โhigh strikerโ plug
is shown at right. The plug starts with a high velocity and begins to slow
down when it hits the bell after 0.60 s.
Strategy: The average velocity is the distance traveled by the plug
divided by the time (Equation 2-10). Assuming there is no friction, the
time and free fall acceleration (โ9.81 m/s2) can be used to find the change
in velocity (Equation 2-7). The initial velocity can then be determined
from the change in velocity and average velocities by combining
Equations 2-7 and 2-9.
x ๏ญ x0 4.0 ๏ญ 0 m
๏ฝ
๏ฝ 6.7 m/s
t
0.60 s
Solution: 1. (a) Find the average
velocity using Equation 2-10:
vav ๏ฝ
2. (b) Find the change in velocity using Eq. 2-7:
๏v ๏ฝ v ๏ญ v0 ๏ฝ at ๏ฝ ๏จ ๏ญ9.81 m/s2 ๏ฉ ๏จ 0.60 s ๏ฉ ๏ฝ ๏ญ5.9 m/s
3. (c) Solve Equation 2-7 for v0 :
v ๏ฝ v0 ๏ซ at ๏ v0 ๏ฝ v ๏ญ at
4. Solve Equation 2-9 for v:
vav ๏ฝ 12 ๏จ v0 ๏ซ v ๏ฉ ๏ v ๏ฝ 2vav ๏ญ v0
5. Substitute the expression for v into
the equation for v0 :
v0 ๏ฝ ๏จ 2vav ๏ญ v0 ๏ฉ ๏ญ a t
6. Now solve that expression for v0 :
v0 ๏ฝ 12 ๏จ 2vav ๏ญ a t ๏ฉ ๏ฝ 12 ๏ฉ๏ซ 2 ๏จ 6.7 m/s ๏ฉ ๏ญ ๏จ ๏ญ9.81 m/s 2 ๏ฉ ๏จ 0.60 s ๏ฉ ๏น๏ป
v0 ๏ฝ 9.6 m/s
Insight: There are several other ways of finding these speeds, including graphical analysis. Try measuring the slope of
the graph at the launch point and the point at which the plug hits the bell to find the initial and final speeds.
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portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
2 โ 34
James S. Walker, Physics, 5th Edition
Chapter 2: One-Dimensional Kinematics
88. Picture the Problem: Chestnut A is dropped from rest.
When it has fallen 2.5 m, chestnut B is thrown downward
with an initial speed vB,0. Both nuts land at the same time
after falling 10.0 m.
Nut B
thrown
branch
vB,0 = ?
2.5 m
Strategy: First find the time it takes for nut A to fall 2.5 m
using the equation of motion for position as a function of
time and acceleration (Equation 2-11). Also find the time
required for nut A to fall the entire 10.0 m. Subtract the
first time from the second to find the time interval over
which nut B must reach the ground in order to land at the
same instant as nut A. Then use Equation 2-11 again to
find the initial velocity vB,0 required in order for nut B to
reach the ground in that time.
Nut A
10.0 m
Both land
ground
2 ๏จ 2.5 m ๏ฉ
2๏ x
๏ฝ
๏ฝ 0.714 s
g
9.81 m/s 2
Solution: 1. Find the time it takes for nut A to fall 2.5 m
by solving Equation 2-11 for t and setting vA,0 = 0.
tA,1 ๏ฝ
2. Find the time it takes for nut A to fall the entire
10.0 m:
tA,total ๏ฝ
3. Subtract the times to find the time over
which nut B must reach the ground:
tB,total ๏ฝ tA,total ๏ญ tA,1 ๏ฝ 1.428 ๏ญ 0.714 s ๏ฝ 0.714 s
vB,0 ๏ฝ
4. Solve Equation 2-11 for vB,0:
2 ๏จ10.0 m ๏ฉ
2๏ x
๏ฝ
๏ฝ 1.428 s
g
9.81 m/s 2
2
๏ x ๏ญ 12 g tB,total
๏ฝ
tB,total
10.0 m ๏ญ 12 ๏จ 9.81 m/s 2 ๏ฉ ๏จ 0.714 s ๏ฉ
2
0.714 s
vB,0 ๏ฝ 10.5 m/s ๏ 11 m/s
Insight: In this problem we kept an additional significant figure than is warranted in steps 1, 2, and 3 in an attempt to
get a more accurate answer in step 4. However, if you choose not to do so, differences in rounding will lead to an
answer of 10 m/s. The specified 2.5 m drop distance for nut A limits the answer to two significant digits, and because
the answer is right between 10 and 11 m/s, it could correctly go either way.
89. Picture the Problem: A rock accelerates from rest straight downward and lands on the surface of the Moon.
Strategy: Employ the relationship between acceleration, displacement, and velocity (Equation 2-12) to find the final
velocity.
Solution: Solve Equation 2-12 for velocity v:
v ๏ฝ v02 ๏ซ 2a๏x ๏ฝ 02 ๏ซ 2 ๏จ1.62 m/s 2 ๏ฉ ๏จ1.25 m ๏ฉ ๏ฝ 2.01 m/s
Insight: On Earth the rock would be traveling 4.95 m/s, but the weaker gravity on the Moon accelerates the rock only
about one-sixth as much as would the Earthโs gravity.
90. Picture the Problem: An elevator in the Taipei 101 skyscraper accelerates to its maximum speed.
Strategy: Because time information is neither given nor requested, the time-free equation for velocity in terms of
displacement (Equation 2-12) is the best choice for finding the displacement.
v 2 ๏ญ v02 ๏จ16 m/s ๏ฉ ๏ญ 0
๏x ๏ฝ
๏ฝ
๏ฝ 116 m
2a
2 ๏จ1.1 m/s 2 ๏ฉ
2
Solution: Solve Equation 2-12 for displacement:
Insight: The observatory elevators in Taipei 101 were the worldโs fastest when installed, whisking passengers from the
fifth floor to the 89th-floor observatory, a distance of 369 m, in only 37 seconds. The 16.83 m/s maximum speed is
equivalent to 37.7 mi/h. The passengers are treated to a memorable ride!
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portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
2 โ 35
James S. Walker, Physics, 5th Edition
Chapter 2: One-Dimensional Kinematics
91. Picture the Problem: Water pouring through an ancient Strait of Gibraltar accelerates downward and impacts the water
surface below.
Strategy: Employ the relationship between acceleration, displacement, and velocity (Equation 2-12) to find the height
from which the water must fall so that its final velocity just before landing is 340 m/s.
2
v 2 ๏ญ v02 ๏จ 340 m/s ๏ฉ ๏ญ 0
๏ฝ
๏ฝ 5900 m ๏ฝ 5.9 km
2g
2 ๏จ 9.81 m/s 2 ๏ฉ
2
Solution: Solve Equation 2-12 for velocity ๏x:
๏x ๏ฝ
Insight: This height corresponds to 3.7 miles or over 19,000 feet! With air resistance, however, an even higher altitude
would be required to obtain speeds this great.
92. Picture the Problem: A juggler throws a ball straight upward, it briefly comes to rest, and falls downward, returning to
the jugglerโs hand.
Strategy: By symmetry the total time of flight is exactly twice the amount of time elapsed as the ball falls from rest
from its maximum height. Use this observation, together with the equation for position as a function of time (Equation
2-11) to find the maximum height of the ball above the jugglerโs hand.
Solution: 1. Solve Equation 2-11 for x0 , assuming that
x ๏ฝ x0 ๏ซ v0 t ๏ซ 12 a t 2
the final position x ๏ฝ 0 and initial speed v0 ๏ฝ 0 :
0 ๏ฝ x0 ๏ซ 0 ๏ซ 12 ๏จ ๏ญ g ๏ฉ t 2
2. Substitute values to find the maximum height:
x0 ๏ฝ 12 ๏จ 9.81 m/s2 ๏ฉ ๏จ 12 ๏ด 0.98 s ๏ฉ ๏ฝ 1.2 m
๏ x0 ๏ฝ 12 g t 2
2
Insight: You can show that the ball left the jugglerโs hand with an upward velocity of 4.8 m/s, or about 11 mi/h.
93. Picture the Problem: Ball A is dropped from rest at the edge of a roof, and at the same instant ball B is thrown upward
from the ground with an initial velocity v0 sufficient to reach the original location of ball A.
Strategy: Use an understanding of velocity and acceleration to answer the conceptual question. Let upward be the
positive direction.
Solution: 1. The velocity of ball A is negative because it is falling downward.
2. The acceleration of ball A is negative because gravity acts in the downward direction.
3. The velocity of ball B is positive because it is traveling upward.
4. The acceleration of ball B is negative because gravity acts in the downward direction.
Insight: Acceleration is the rate of change of velocity, so acceleration can be zero when the velocity has a large
magnitude (for example, a car traveling along a highway at constant speed), and the velocity can be zero when the
acceleration has a large magnitude (for example, a ball at the top of its vertical flight). The acceleration of ball B is
always downward, even when its velocity is upward.
94. Picture the Problem: Two balls are released simultaneously. Ball A is
dropped from rest but ball B is thrown upward with an initial velocity v0 .
Strategy: Use a correct interpretation of motion graphs to answer the
conceptual questions. Recall that the slope of a velocity-versus-time graph
is the acceleration.
Solution: 1. (a) The speed of ball A starts at zero and then increases
linearly with a slope of 9.81 m/s2. The graph that corresponds to that
description is plot 3.
2. (b) The speed of ball B starts at v0 and then decreases linearly with a slope of โ9.81 m/s2, equal in magnitude but
opposite in direction to the slope of ball Aโs plot. The graph that corresponds to that description is plot 2.
Insight: Even if ball B were fired upward at an extremely high speed, its velocity-versus-time graph would still be
linear with a slope of โ9.81 m/s2, but the line would begin very high on the speed axis of the graph.
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portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
2 โ 36
James S. Walker, Physics, 5th Edition
Chapter 2: One-Dimensional Kinematics
95. Picture the Problem: A plot of position vs. time yields information about
the average velocity of an object.
Strategy: Use a correct interpretation of position-time plots to answer the
conceptual questions. Recall that the slope of a position-versus-time graph
is the velocity.
Solution: 1. (a) The average speed of the object is the magnitude of the
slope of a line between the two points. The line between points 1 and 2
(dark solid line) has a steeper slope than the line between points 1 and 3
(light solid line). Therefore, the average speed for the time interval
between points 1 and 2 is greater than the average speed for the time
interval between points 1 and 3.
2. (b) The average velocity of the object is the slope of a line between the two points. The line between points 2 and 4
(dark dashed line) has a smaller slope than the line between points 3 and 4 (light dashed line). Therefore, the average
velocity for the time interval between points 2 and 4 is less than the average velocity for the time interval between
points 3 and 4.
Insight: The negative slopes of the two solid lines indicate the velocity of the object is negative for those time intervals.
However, the question in part (a) asked about the speed, not the velocity, hence only the magnitude of the slopes was
considered.
96. Picture the Problem: A package falls straight downward, accelerating for 2.2 seconds before impacting air bags.
Strategy: Find the distance the package will fall from rest in 2.2 seconds by using Equation 2-11. Use the known
acceleration and time to find the velocity of the package just before impact by using Equation 2-7. Finally, use the
known initial and final velocities, together with the distance over which the package comes to rest when in contact with
the air bags, to find the stopping acceleration using Equation 2-12.
Solution: 1. (a) Find the distance the package
falls from rest in 2.2 s using Equation 2-11:
๏ x ๏ฝ v0 t ๏ซ 12 g t 2 ๏ฝ 0 ๏ซ 12 ๏จ 9.81 m/s2 ๏ฉ ๏จ 2.2 s ๏ฉ ๏ฝ 24 m
2. (b) Find the velocity just
before impact using Equation 2-7:
vland ๏ฝ v0 ๏ซ g t ๏ฝ 0 ๏ซ ๏จ 9.81 m/s2 ๏ฉ ๏จ 2.2 s ๏ฉ ๏ฝ 22 m/s ๏ฝ 48 mi/h!
3. (c) Solve Equation 2-12 for a:
a๏ฝ
2
v 2 ๏ญ v02 0 ๏ญ ๏จ 22 m/s ๏ฉ
๏ฝ
๏ฝ ๏ญ320 m/s 2 ๏ฝ ๏ญ33g
2๏ x
2 ๏จ 0.75 m ๏ฉ
2
2
Insight: Increasing the stopping distance will decrease the stopping acceleration. We will return to this idea when we
discuss impulse and momentum in Chapter 9.
Copyright ยฉ 2017 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No
portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
2 โ 37
James S. Walker, Physics, 5th Edition
Chapter 2: One-Dimensional Kinematics
97. Picture the Problem: A plot of velocity vs. time yields information about
the acceleration of an object.
Strategy: Use a correct interpretation of motion graphs to answer the
conceptual questions. Recall that the slope of a velocity-versus-time graph
is the acceleration.
Solution: 1. (a) The acceleration of the object is the slope of the velocityv ๏ญ v0 4.5 ๏ญ 0.5 m/s
versus-time graph: a ๏ฝ
๏ฝ
๏ฝ 4.0 m/s 2
t ๏ญ t0
1.0 ๏ญ 0 s
2. (b) The displacement is the area under the velocityversus-time curve:
๏ x ๏ฝ area of bottom rectangle ๏ซ area of triangle
3. The final position is the initial position plus the
displacement:
x ๏ฝ x0 ๏ซ ๏ x ๏ฝ 12.0 m ๏ซ 2.5 m ๏ฝ 14.5 m
4. (c) Use the known acceleration, initial velocity, and
initial position to find the final position at t = 5.00 s:
x ๏ฝ x0 ๏ซ v0 t ๏ซ 12 a t 2
๏ฝ ๏จ1.0 s ๏ฉ๏จ 0.50 m/s ๏ฉ ๏ซ 12 ๏จ1.0 s ๏ฉ๏จ 4.0 m/s ๏ฉ ๏ฝ 2.5 m
๏ฝ ๏จ12.0 m ๏ฉ ๏ซ ๏จ 0.50 m/s ๏ฉ๏จ 5.00 s ๏ฉ ๏ซ 12 ๏จ 4.0 m/s 2 ๏ฉ ๏จ 5.00 s ๏ฉ
2
x ๏ฝ 64.5 m
Insight: Equation 2-11 can also be used to find the final position in part (b), instead of determining the area under the
velocity-versus-time graph: x ๏ฝ x0 ๏ซ v0 t ๏ซ 12 a t 2 ๏ฝ ๏จ12.0 m ๏ฉ ๏ซ ๏จ 0.50 m/s ๏ฉ๏จ1.00 s ๏ฉ ๏ซ 12 ๏จ 4.0 m/s2 ๏ฉ ๏จ1.00 s ๏ฉ ๏ฝ 14.5 m.
2
98. Picture the Problem: A golf ball rolls in a straight line, decreasing its speed at a constant rate until it comes to rest.
Strategy: You could find the (negative) acceleration by using Equation 2-12 and the known initial and final velocities
and the distance traveled. Then employ Equation 2-12 again using the same acceleration, but solving for the v0 required
to go the longer distance. Instead, weโll present a way to calculate the same answer using a ratio, which will also be
useful to calculate the initial speed needed to make the putt over the new 6.00-ft distance.
v 2 ๏ญ 2 a ๏ xmake
02 ๏ญ 2 a ๏ xmake
๏ xmake
๏ xmiss
Solution: 1. (a) Calculate the ratio of initial
velocities based upon Equation 2-12:
vmake,0
2. Now solve for vmake,0 , the initial speed
vmake,0 ๏ฝ vmiss,0
๏ xmake
23.5 ft
๏ฝ ๏จ1.54 m/s ๏ฉ
๏ฝ 1.77 m/s
๏ xmiss
23.5 ๏ญ 6.00 ft
vnew,0 ๏ฝ vmiss,0
๏ xnew
6.00 ft
๏ฝ ๏จ1.54 m/s ๏ฉ
๏ฝ 0.902 m/s
๏ xmiss
23.5 ๏ญ 6.00 ft
needed to make the 23.5-ft putt:
3. (b) Employ the same ratio to find the
initial speed for the new 6.00-ft putt:
vmiss,0
๏ฝ
v ๏ญ 2 a ๏ xmiss
2
๏ฝ
0 ๏ญ 2 a ๏ xmiss
2
๏ฝ
Insight: Calculating ratios can often be a convenient and simple way to solve a problem. In this case a three-step
solution became two steps when we calculated the ratio, and furthermore we never needed to convert feet to meters
because the units cancel out in the ratio. Learning to calculate ratios in this manner is a valuable skill in physics.
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portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
2 โ 38
James S. Walker, Physics, 5th Edition
Chapter 2: One-Dimensional Kinematics
99. Picture the Problem: After its release by a glaucous-winged gull, a shell rises straight upward, slows down, and
momentarily comes to rest before falling straight downward again.
Strategy: Find the extra altitude attained by the shell due to its upward initial velocity upon release, and add that value
to 12.5 m to find the maximum height it reaches above ground. The time-free equation for velocity in terms of
displacement (Equation 2-12) can be employed for this purpose. The time the shell spends going up and the time it
spends going down can each be found from the known heights and speeds (Equations 2-7 and 2-11). Then the speed
upon landing can be determined from the known time it spends falling (Equation 2-7). Let upward be the positive
direction throughout the solution to this problem.
Solution: 1. (a) The motion of the shell is influenced only by gravity once it has been released by the gull. Therefore its
acceleration will be 9.81 m/s2 downward from the moment it is released, even though it is moving upward at the
release.
02 ๏ญ ๏จ 5.20 m/s ๏ฉ
v 2 ๏ญ v02
xmax ๏ฝ 12.5 m ๏ซ
๏ฝ 12.5 m ๏ซ
๏ญ2 g
๏ญ2 ๏จ 9.81 m/s 2 ๏ฉ
2
2. (b) Use Equation 2-12, setting the
final speed v = 0, to find the extra altitude
gained by the shell due to its initial upward
speed, and add it to the 12.5 m:
xmax ๏ฝ 12.5 m ๏ซ 1.38 m ๏ฝ 13.9 m
3. (c) The time the shell travels upward is the time it
takes gravity to bring the speed to zero (Equation 2-7):
t๏ฝ
v ๏ญ v0 0 ๏ญ 5.2 m/s
๏ฝ
๏ฝ 0.53 s
๏ญg
๏ญ9.81 m/s 2
x ๏ฝ x0 ๏ซ v0t ๏ญ 12 gt 2 ๏ 0 ๏ฝ x0 ๏ซ 0 ๏ญ 12 gt 2
4. The time the shell travels down is governed by
the distance and the acceleration (Equation 2-11):
t๏ฝ
2 x0
๏ฝ
g
2 ๏จ13.9 m ๏ฉ
9.81 m/s 2
๏ฝ 1.68 s
ttotal ๏ฝ tup ๏ซ tdown ๏ฝ 0.53 ๏ซ 1.68 s ๏ฝ 2.21 s
5. The total time of flight is the sum:
v ๏ฝ v0 ๏ญ g t ๏ฝ 0 ๏ญ ๏จ 9.81 m/s 2 ๏ฉ ๏จ1.68 s ๏ฉ ๏ฝ ๏ญ16.5 m/s
6. (d) The speed of the shell upon impact
is given by the acceleration of gravity and
the fall time (Equation 2-7):
v ๏ฝ 16.5 m/s
Insight: There are a variety of other ways to solve this problem. For instance, it is possible to find the final velocity of
16.5 m/s in part (d) by using Equation 2-12 with v0 ๏ฝ 5.2 m/s and ๏ x ๏ฝ ๏ญ12.5 m without using any time information.
Try it for yourself!
100. Picture the Problem: Liquid from a syringe squirts straight upward, slows down, and momentarily comes to rest before
falling straight downward again.
Strategy: Find the time of flight by exploiting the symmetry of the situation. If it takes time t for gravity to slow the
liquid drops down from their initial speed v0 to zero, it will take the same amount of time to accelerate them back to the
same speed. They therefore return to the needle tip at the same speed v0 with which they were squirted. Use this fact
together with Equation 2-7 to find the time of flight. The maximum height the drops achieve is related to the square of
v0, as indicated by Equation 2-12.
v ๏ญ v0 ๏จ ๏ญv0 ๏ฉ ๏ญ v0 2v0 2 ๏จ1.5 m/s ๏ฉ
๏ฝ
๏ฝ
๏ฝ
๏ฝ 0.31 s
๏ญg
๏ญg
g
9.81 m/s 2
Solution: 1. (a) Calculate he time of flight
for v0 ๏ฝ 1.5 m/s , using Equation 2-7:
t๏ฝ
2. (b) Calculate the maximum height for
v0 ๏ฝ 1.5 m/s , using Equation 2-12:
v 2 ๏ญ v02 02 ๏ญ v02 v02
๏จ1.5 m/s ๏ฉ
๏x ๏ฝ
๏ฝ
๏ฝ
๏ฝ
๏ฝ 0.11 m
๏ญ2 g
๏ญ2 g
2 g 2 ๏จ 9.81 m/s 2 ๏ฉ
2
Insight: The symmetry of the motion of a freely falling object can often be a useful tool for solving problems quickly.
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portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
2 โ 39
James S. Walker, Physics, 5th Edition
Chapter 2: One-Dimensional Kinematics
101. Picture the Problem: The trajectories of a hot-air balloon and a
camera are shown at right. The balloon rises at a steady rate
while the cameraโs speed is continually slowing down under the
influence of gravity. The camera is caught when the two
trajectories meet.
Strategy: The equation of motion for position as a function of
time (Equation 2-10) can be used to describe the balloon, while
the equation for position as a function of time and acceleration
(Equation 2-11) can be used to describe the cameraโs motion.
Set these two equations equal to each other to find the time at
which the camera is caught. Then find the height of the balloon
at the instant the camera is caught.
Solution: 1. Write Equation 2-10 for the balloon:
xb ๏ฝ xb,0 ๏ซ vb t
2. Write Equation 2-11 for the camera:
xc ๏ฝ 0 ๏ซ vc,0 t ๏ญ 12 g t 2
3. Set xb ๏ฝ xc and solve for t:
xb,0 ๏ซ vb t ๏ฝ vc,0 t ๏ญ 12 g t 2
0 ๏ฝ ๏ญ xb,0 ๏ซ ๏จ vc,0 ๏ญ vb ๏ฉ t ๏ญ 12 g t 2
0 ๏ฝ 2.5 m ๏ญ ๏จ13 ๏ญ 2.0 m/s ๏ฉ t ๏ซ 12 ๏จ 9.81 m/s 2 ๏ฉ t 2
4. Multiply by โ1 and substitute the numerical values:
0 ๏ฝ 2.5 ๏ญ 11t ๏ซ 4.9 t 2
2
๏ญb ๏ฑ b 2 ๏ญ 4ac ๏ซ11 ๏ฑ 11 ๏ญ 4 ๏จ 4.9 ๏ฉ๏จ 2.5 ๏ฉ
๏ฝ
2a
9.8
t ๏ฝ 0.26 or 2.0 s
5. Apply the quadratic formula and solve for t. The
larger root corresponds to the time when the camera
would pass the balloon a second time, on its way
down back to the ground.
t๏ฝ
xb ๏ฝ xb,0 ๏ซ vb t ๏ฝ 2.5 m ๏ซ ๏จ 2.0 m/s ๏ฉ๏จ 0.26 s ๏ฉ ๏ฝ 3.0 m
6. Find the height of the balloon at that time:
Insight: If the passenger misses the camera the first time, she has another shot at it after 2.0 s (from the time it is
thrown) when the camera is on its way back toward the ground. That is the meaning of the second solution for t.
102. Picture the Problem: The height-versus-time plot of a rock on a
distant planet is shown at right. The rock starts with a high
velocity upward, slows down and momentarily comes to rest
after about 4.0 seconds of flight, and then falls straight down
and lands at about 8.0 seconds.
Strategy: The equation of motion for position as a function of
time and acceleration (Equation 2-11) can be used to find the
acceleration from the second half of the trajectory, where the
rock falls 30 m from rest and lands 4.0 seconds later. Once
acceleration is known, the initial velocity can be determined
from Equation 2-7. Let upward be the positive direction.
Solution: 1. (a) Solve Equation 2-11 for acceleration, assuming v0 ๏ฝ 0 at the peak of its flight
and the rock falls 30 m in 4.0 s:
a๏ฝ
2๏ x 2 ๏จ ๏ญ30 m ๏ฉ
๏ฝ
๏ฝ ๏ญ3.8 m/s 2
2
t2
๏จ 4.0 s ๏ฉ
๏ a ๏ฝ 3.8 m/s 2
v ๏ฝ v0 ๏ซ a t
2. (b) Find the initial velocity using Eq. 2-7,
concentrating on the first half of the flight that
ends with v = 0 at the peak:
v0 ๏ฝ v ๏ญ a t ๏ฝ 0 ๏ญ ๏จ ๏ญ3.8 m/s 2 ๏ฉ ๏จ 4.0 s ๏ฉ ๏ฝ 15 m/s
Insight: There are several other ways of finding the answers, including graphical analysis. Try measuring the slope of
the graph at the launch point and the point at which the rock lands to find the initial and final velocities. Those values
(about ยฑ15 m/s) can then be used to find the acceleration.
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portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
2 โ 40
James S. Walker, Physics, 5th Edition
Chapter 2: One-Dimensional Kinematics
103. Picture the Problem: A squid emits a jet of water, propelling itself forward with constant acceleration, then coasts to
rest with constant (negative) acceleration.
Strategy: The magnitude of the acceleration can be determined from the position as a function of time equation
(Equation 2-11) and the given information. During the first part of the squidโs motion the initial velocity is zero, and
during the second part the final velocity is zero.
x1 ๏ฝ 0 ๏ซ 0 ๏ซ 12 a1 t12
Solution: 1. (a) Find the acceleration during the
first part of the squidโs motion, noting that
x0 ๏ฝ v0 ๏ฝ 0 :
a1 ๏ฝ
2. (b) Find the squidโs velocity at the end of the
first part of its motion:
v ๏ฝ v0 ๏ซ a t1 ๏ฝ 0 ๏ซ ๏จ12.4 m/s2 ๏ฉ ๏จ 0.170 s ๏ฉ ๏ฝ 2.11 m/s
3. The time elapsed during the second part of the
squidโs motion is found by subtraction:
t2 ๏ฝ ttotal ๏ญ t1 ๏ฝ 0.400 ๏ญ 0.170 s ๏ฝ 0.230 s
4. The distance traveled during the second part of
the squidโs motion is found by subtraction:
x2 ๏ฝ xtotal ๏ญ x1 ๏ฝ 0.421 ๏ญ 0.179 m ๏ฝ 0.242 m
5. Calculate the squidโs acceleration during the
second part of its motion:
x2 ๏ฝ 0 ๏ซ v t2 ๏ซ 12 a2 t22
a2 ๏ฝ
2 x1 2 ๏จ 0.179 m ๏ฉ
๏ฝ
๏ฝ 12.4 m/s 2
2
t12
๏จ 0.170 s ๏ฉ
2 ๏จ x2 ๏ญ v t2 ๏ฉ
t22
2 ๏ฉ๏จ 0.242 m ๏ฉ ๏ญ ๏จ 2.11 m/s ๏ฉ๏จ 0.230 s ๏ฉ ๏น๏ป
๏ฝ ๏ซ
2
๏จ 0.230 s ๏ฉ
a2 ๏ฝ ๏ญ 9.20 m/s 2
Insight: Notice that the answer to part (b) can also be determined without finding the squidโs velocity. Instead, work
backward and pretend the squid starts from rest and covers 0.242 m in 0.230 s. Then
a2 ๏ฝ 2 x2 t22 ๏ฝ 2 ๏จ 0.242 m ๏ฉ ๏จ 0.230 s ๏ฉ ๏ฝ 9.15 m/s 2 , which is the same result to within rounding error, although you
must recognize that the acceleration is negative, not positive.
2
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portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
2 โ 41
James S. Walker, Physics, 5th Edition
Chapter 2: One-Dimensional Kinematics
104. Picture the Problem: A ball falls straight downward from rest at an initial height h.
Strategy: The problem requires that the time to fall the final 3/4 h from rest is 1.00 s. Find the
velocity v1 at ยพ h above the ground using Equation 2-12. Use Equation 2-11 along with that initial
velocity and the time elapsed to determine h. Then the total time of fall can be found using
Equation 2-11 again, this time with an initial velocity of zero.
Solution: 1. (a) Find the velocity v1 of the
ball after falling a distance ยผ h:
2. Now insert that velocity as the initial
velocity for the remaining portion of the
fall into Equation 2-11:
h
ยพh
v12 ๏ฝ 02 ๏ซ 2 g ๏ x ๏ฝ 2 g ๏จ 14 h ๏ฉ ๏ v1 ๏ฝ
1
2
gh
๏ x ๏ฝ v1 t ๏ซ 12 g t 2
3
4
h๏ฝ
๏จ
1
2
๏ฉ
gh t ๏ซ 12 g t 2
3. The time t is 1.00 s as given in the problem
statement. Rearrange the above equation and
square both sides to get a quadratic equation:
3
4
9
16
h ๏ญ 12 g t 2 ๏ฝ
๏จ
1
2
h ๏ญ 2 ๏จ g t ๏ฉ ๏จ h๏ฉ ๏ซ g t ๏ฝ g h t
2
2
1
2
9
16
3
4
2 4
1
4
1
2
๏ฉ
gh t
2
h 2 ๏ญ ๏จ 54 g t 2 ๏ฉ h ๏ซ 14 g 2t 4 ๏ฝ 0
h 2 ๏ญ ๏จ 209 g t 2 ๏ฉ h ๏ซ 94 g 2 t 4 ๏ฝ 0
h 2 ๏ญ 209 ๏จ 9.81 m/s 2 ๏ฉ ๏จ1.00 s ๏ฉ h ๏ซ 94 ๏จ 9.81 m/s 2 ๏ฉ ๏จ1.00 s ๏ฉ ๏ฝ 0
2
2
4
h 2 ๏ญ 21.8h ๏ซ 42.8 ๏ฝ 0
4. Now apply the quadratic formula for h:
๏ญb ๏ฑ b 2 ๏ญ 4ac 21.8 ๏ฑ
h๏ฝ
๏ฝ
2a
5. (b) Use Equation 2-11 again to find
the total time of fall:
t๏ฝ
2h
๏ฝ
g
2 ๏จ19.6 m ๏ฉ
9.81 m/s 2
๏จ 21.8๏ฉ ๏ญ 4 ๏จ1๏ฉ๏จ 42.8๏ฉ
๏ฝ 2.18, 19.6 m
2 ๏จ1๏ฉ
2
๏ฝ 2.00 s
Insight: The first root in step 4 (2.18 m) is thrown out because the total fall time from that height would be less than
1.00 s, but the ball is supposed to be in the air for longer than 1.00 s. Notice it takes half the total flight time to fall the
first quarter of the fall distance, and half to fall the final three quarters.
105. Picture the Problem: A ski glove falls straight downward from rest, accelerates to a maximum speed under the
influence of gravity, then decelerates due to its interaction with the snow before coming to rest at a depth d below the
surface of the snow.
Strategy: We can find the maximum speed of the glove from its initial height and the acceleration of gravity by using
Equation 2-12. The same equation can be applied again, this time with a zero final speed instead of zero initial speed, to
find the acceleration caused by the snow. Let downward be the positive direction.
Solution: 1. (a) Solve Equation 2-12
for v, assuming v0 ๏ฝ 0 :
v ๏ฝ 02 ๏ซ 2 gh ๏ฝ
2 gh
2. (b) Use Equation 2-12 to find
the acceleration caused by the snow:
02 ๏ฝ v02 ๏ซ 2ad ๏
๏ญ 2ad ๏ฝ
๏จ 2 gh ๏ฉ
2
๏ a๏ฝ ๏ญ
h
g
d
3. The negative sign on the acceleration means the glove is accelerated upward during its interaction with the snow.
Insight: In Chapter 5 we will analyze the motion of objects like this glove in terms of force vectors. This motion can
also be explained in terms of energy using the tools introduced in Chapters 7 and 8.
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portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
2 โ 42
James S. Walker, Physics, 5th Edition
Chapter 2: One-Dimensional Kinematics
106. Picture the Problem: A ball rises straight upward, passes a power line, momentarily comes to rest, and falls back to
Earth again, passing the power line a second time on its way down.
Strategy: The ball will reach the peak of its flight at a time directly between the times it passes the power line. The
time to reach the peak of flight can be used to find the initial velocity using Equation 2-7, and the initial velocity can
then be used to find the height of the power lines using Equation 2-11.
Solution: 1. Find the time at which
the ball reaches its maximum altitude:
2. Find the initial velocity using Equation 2-7:
3. Find the height of the power
line using Equation 2-11:
tpeak ๏ฝ tline up ๏ซ 12 ๏จ tline down ๏ญ tline up ๏ฉ ๏ฝ 0.75 s ๏ซ 12 ๏จ1.5 ๏ญ 0.75 s ๏ฉ
tpeak ๏ฝ 1.1 s
0 ๏ฝ v0 ๏ญ g tpeak ๏ v0 ๏ฝ ๏จ 9.81 m/s2 ๏ฉ ๏จ1.1 s ๏ฉ ๏ฝ 11 m/s
2
x ๏ฝ 0 ๏ซ v0 tline up ๏ญ 12 g tline
up
x ๏ฝ ๏จ11 m/s ๏ฉ๏จ 0.75 s ๏ฉ ๏ญ 12 ๏จ 9.81 m/s 2 ๏ฉ ๏จ 0.75 s ๏ฉ ๏ฝ 5.5 m
2
Insight: As is often the case, there are several other ways to solve this problem. Try setting the heights at 0.75 s and
1.5 s equal to each other and solving for v0. Can you think of yet another way?
107. Picture the Problem: A ball appears at the bottom edge of the window, rising straight
upward with initial speed v0. It travels upward, disappearing beyond the top edge of the
window, comes to rest momentarily, and then falls straight downward, reappearing some
time later at the top edge of the window. In the drawing at right the motion of the ball is
offset horizontally for clarity.
Strategy: Let t = 0 correspond to the instant the ball first appears at the bottom edge of the
window with speed v0. Write the equation of position as a function of time and acceleration (Equation 2-11) for when the ball is at the top edge (position 2) in order to find v0.
Use v0 to find the time to go from position 1 to the peak of the flight (Equation 2-7).
Subtract 0.25 s from that time to find the time to go from position 2 to the peak of the
flight. The time elapsed between positions 2 and 3 is twice the time to go from position 2
to the peak of the flight. The time from position 2 to the peak can be used to find h from
Equation 2-11.
Solution: 1. (a) Write Equation 2-11
for positions 1 and 2, and solve for v0:
h
2
3
d
v0
1
d ๏ฝ v0 t2 ๏ญ 12 g t22
d ๏ซ 12 g t22 1.05 m ๏ซ 2 ๏จ 9.81 m/s ๏ฉ ๏จ 0.25 s ๏ฉ
v0 ๏ฝ
๏ฝ
๏ฝ 5.4 m/s
t2
0.25 s
1
2
2
0 ๏ญ v0
5.4 m/s
๏ฝ
๏ฝ 0.55 s
๏ญg
9.81 m/s 2
2. Find the time to go from position 1 to
the peak of the flight using Equation 2-7:
๏t1, p ๏ฝ
3. Subtract 0.25 s to find the time to go
from position 2 to the peak of the flight:
๏t2, p ๏ฝ ๏t1, p ๏ญ ๏t1,2 ๏ฝ 0.55 ๏ญ 0.25 s ๏ฝ 0.30 s
4. The time to reappear is twice this time:
๏t2,3 ๏ฝ 2๏t2, p ๏ฝ 2 ๏จ 0.30 s ๏ฉ ๏ฝ 0.60 s
5. (b) The height h can be found from ๏t2, p
0 ๏ฝ h ๏ซ 0 ๏ญ 12 g ๏t2,2 p
and Equation 2-11, by considering the ball
dropping from rest at the peak to position 3:
h ๏ฝ 12 ๏จ 9.81 m/s 2 ๏ฉ ๏จ 0.30 s ๏ฉ ๏ฝ 0.44 m
2
Insight: As usual there are other ways to solve this problem. Try finding the velocity at position 2 and use it together
with the acceleration of gravity and the average velocity from position 2 to the peak to find ๏t2,3 and h.
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2 โ 43
James S. Walker, Physics, 5th Edition
Chapter 2: One-Dimensional Kinematics
108. Picture the Problem: The lunar lander falls straight downward, accelerating over a distance of 4.30 ft before impacting
the lunar surface.
Strategy: Use the given acceleration and distance and the time-free equation of motion (Equation 2-12) to find the
velocity of the lander just before impact. Use the known initial and final velocities, together with the distance of the
fall, to find the time elapsed using Equation 2-10.
Solution: 1. Find the velocity just
before impact using Equation 2-12:
vland ๏ฝ v02 ๏ซ 2a๏ x
2. Solve Equation 2-10 for tfall:
tfall ๏ฝ 1
๏ฝ
๏จ 0.500 ft/s ๏ฉ ๏ซ 2 ๏จ1.62 m/s2 ๏ด 3.28 ft/m ๏ฉ ๏จ 4.30 ft ๏ฉ ๏ฝ 6.78 ft/s
2
๏ xfall
4.30 ft
๏ฝ1
๏ฝ 1.18 s
v
๏ซ
v
0.500
๏ซ 6.78 ft/s ๏ฉ
๏จ
๏ฉ
๏จ
0
land
2
2
Insight: An alternative strategy would be to solve Equation 2-11 as a quadratic equation in t. Assuming the lander feet
had little in the way of shock absorbers, the lander came to rest in a distance given by the amount the lunar dust
compacted underneath the feet. Supposing it was about 2 cm, the astronauts experienced a brief deceleration of
106 m/s2 = 11g! Bam!
109. Picture the Problem: The lunar lander falls straight downward, accelerating over a distance of 4.30 ft before impacting
the lunar surface.
Strategy: Use the given acceleration and distance and the time-free equation of motion (Equation 2-12) to find the
velocity of the lander just before impact.
Solution: Find the velocity just
before impact using Equation 2-12:
vland ๏ฝ v02 ๏ซ 2a๏ x
๏ฝ
๏จ 0.500 ft/s ๏ฉ ๏ซ 2 ๏จ1.62 m/s2 ๏ด 3.28 ft/m ๏ฉ ๏จ 4.30 ft ๏ฉ ๏ฝ 6.78 ft/s
2
Insight: The initial speed made little difference; if you set v0 ๏ฝ 0 youโll note that vland ๏ฝ 6.76 ft/s.
110. Picture the Problem: The lunar lander falls straight downward, accelerating over a distance of 4.30 ft before impacting
the lunar surface.
Strategy: The lander has an initial downward velocity and accelerates downward at a constant rate. Use the knowledge
that the velocity-versus-time graph is a straight line for constant acceleration to determine which graph is the
appropriate one.
Solution: Graph B is the only one that depicts the speed increasing linearly with time.
Insight: Graph D would be an appropriate depiction of the altitude versus time graph.
111. Picture the Problem: We imagine that the astronauts increase the upward thrust, giving the lunar lander a small upward
acceleration.
Strategy: The lander has an initial downward velocity and accelerates upward at a constant rate. This means the
landerโs speed would decrease at a constant rate. Use the knowledge that the velocity-versus-time graph is a straight line
for constant acceleration to determine which graph is the appropriate one.
Solution: Plot C is the only one that depicts the speed decreasing linearly with time.
Insight: The altitude-versus-time graph in this case would curve upward much like plot A but would have an initially
negative slope like plot D.
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portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
2 โ 44
James S. Walker, Physics, 5th Edition
Chapter 2: One-Dimensional Kinematics
112. Picture the Problem: The trajectories of the speeder and police
car are shown at right. The speeder moves at a constant velocity
while the police car has a constant acceleration, except the police
car is delayed in time from when the speeder passes it at x = 0.
Strategy: The equation of motion for position as a function of time
and velocity (Equation 2-10) can be used to describe the speeder,
while the equation for position as a function of time and
acceleration (Equation 2-11) can be used to describe the police
carโs motion. Set these two equations equal to each other and
solve the resulting equation to find the speederโs head-start xshs .
Solution: 1. Write Equation 2-10 for
the speeder, with t = 0 corresponding
to the instant it passes the police car:
xs ๏ฝ xshs ๏ซ vs t
2. Write Equation 2-11 for the police car:
xp ๏ฝ 0 ๏ซ 0 ๏ซ 12 ap t 2
3. Set xp ๏ฝ xs and solve for xshs :
1
2
ap t 2 ๏ฝ xshs ๏ซ vs t
xshs ๏ฝ 12 ap t 2 ๏ญ vs t ๏ฝ 12 ๏จ 3.8 m/s 2 ๏ฉ ๏จ15 s ๏ฉ ๏ญ ๏จ 25 m/s ๏ฉ๏จ15 s ๏ฉ
2
xshs ๏ฝ 53 m
Insight: This head start corresponds to about 2.10 seconds (verify for yourself, and/or examine the plot) so the police
officer has to be ready to start the chase very soon after the speeder passes by!
113. Picture the Problem: The trajectories of the speeder and police
car are shown at right. The speeder moves at a constant velocity
while the police car has a constant acceleration.
Strategy: The equation of motion for position as a function of
time and velocity (Equation 2-10) can be used to describe the
speeder, while the equation for position as a function of time and
acceleration (Equation 2-11) can be used to describe the police
carโs motion. Set these two equations equal to each other and
solve the resulting equation for the acceleration of the police car.
Solution: 1. Write Equation 2-10 for
the speeder, with t = 0 corresponding
to the instant it passes the police car:
xs ๏ฝ 0 ๏ซ vs t
2. Write Equation 2-11 for the police car:
xp ๏ฝ 0 ๏ซ 0 ๏ซ 12 ap t 2
3. Set xp ๏ฝ xs and solve for ap :
1
2
ap t 2 ๏ฝ vs t
ap ๏ฝ
2vs 2 ๏จ15 m/s ๏ฉ
๏ฝ
๏ฝ 4.3 m/s 2
t
7.0 s
Insight: A faster acceleration of the police car would allow it to catch the speeder in less than 7.0 s.
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portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
2 โ 45
James S. Walker, Physics, 5th Edition
Chapter 2: One-Dimensional Kinematics
114. Picture the Problem: The trajectory of a bag of sand is shown at right.
After release from the balloon it rises straight up and comes
momentarily to rest before accelerating straight downward and
impacting the ground.
Strategy: Because the initial velocity, acceleration, and altitude are
known, we need only use Equation 2-12 to find the final velocity.
Solution: 1. (a) Because the upward speed of the sandbag is the same, it
will gain the same additional 2 m in altitude as it did in the original
Example 2-12. Therefore the maximum height will be equal to 32 m.
v 2 ๏ฝ v02 ๏ซ 2a๏ x
2. (b) Apply Equation 2-12 to find the final velocity:
v๏ฝ
๏จ 6.5 m/s ๏ฉ ๏ซ 2 ๏จ ๏ญ9.81 m/s2 ๏ฉ ๏จ ๏ญ30.0 m ๏ฉ ๏ฝ 25 m/s
2
Insight: Another way to find the final velocity just before impact is to allow the sandbag to fall from rest a distance of
32 m. Try it!
115. Picture the Problem: A bag of sand has an initial downward velocity when it breaks free from the balloon, and is
accelerated by gravity until it hits the ground.
Strategy: Because the initial velocity, acceleration, and altitude are known, we need only use Equation 2-12 to find the
final velocity. The time can then be found from the average velocity and the distance.
Solution: 1. (a) Apply Equation 2-12 to find the final v:
v 2 ๏ฝ v02 ๏ซ 2a๏ x
๏จ 4.2 m/s ๏ฉ ๏ซ 2 ๏จ ๏ญ9.81 m/s2 ๏ฉ ๏จ ๏ญ35.0 m ๏ฉ ๏ฝ 26.5 m/s
v๏ฝ
2
x ๏ญ x0
0 ๏ญ 35 m
๏ฝ1
๏ฝ 2.3 s
v
๏ซ
v
๏ญ
4.5
๏ญ 26.5 m/s ๏ฉ
๏จ
๏ฉ
๏จ
0
2
2
2. Use Equation 2-10 to find the time:
t๏ฝ 1
3. (b) Apply Equation 2-12 again to find v at x = 15 m:
v 2 ๏ฝ v02 ๏ซ 2a๏ x
v๏ฝ
๏จ 4.2 m/s ๏ฉ ๏ซ 2 ๏จ ๏ญ9.81 m/s2 ๏ฉ ๏จ15 ๏ญ 35 m ๏ฉ ๏ฝ 20 m/s
2
Insight: Another way to find the descent time of the bag of sand is to solve Equation 2-11 using the quadratic formula.
Try it!
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portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
2 โ 46
Chapter 2: One-Dimensional Kinematics
Answers to Conceptual Questions
116. Can you drive your car in such a way that the distance it covers is (a) greater than, (b) equal to, or (c) less than the
magnitude of its displacement? In each case, give an example if your answer is yes, explain why not if your answer is
no.
(a) Yes. If you drive in a complete circle your distance is the circumference of the circle, but your displacement is zero.
(b) Yes. The distance and the magnitude of the displacement are equal if you drive in a straight line. (c) No. Any
deviation from a straight line results in a distance that is greater than the magnitude of the displacement.
117. CE Predict/Explain You drive your car in a straight line at 15 m/s for 10 minutes, then at 25 m/s for another
10 minutes. (a) Is your average speed for the entire trip more than, less than, or equal to 20 m/s? (b) Choose the best
explanation from among the following:
I. More time is required to drive at 15 m/s than at 25 m/s.
II. Less distance is covered at 25 m/s than at 15 m/s.
III. Equal time is spent at 15 m/s and 25 m/s.
Yes. For example, your friends might have backed out of a parking place at some point in the trip, giving a negative
velocity for a short time.
118. In 1992 Zhuang Yong of China set a womenโs Olympic record in the 100-meter freestyle swim with a time of 54.64
seconds. What was her average speed in m/s and mi/h?
No. If you throw a ball upward, for example, you might choose the release point to be y = 0. This doesnโt change the
fact that the initial upward speed is nonzero.
119. A finch rides on the back of a Galapagos tortoise, which walks at the stately pace of 0.060 m/s. After 1.2 minutes the
finch tires of the tortoiseโs slow pace, and takes flight in the same direction for another 1.2 minutes at 12 m/s. What was
the average speed of the finch for this 2.4-minute interval?
Ignoring air resistance, the two gloves have the same acceleration.
Solutions to Problems and Conceptual Exercises
120. In 1992 Zhuang Yong of China set a womenโs Olympic record in the 100-meter freestyle swim with a time of 54.64
seconds. What was her average speed in m/s and mi/h?
Picture the Problem: The swimmer swims in the forward direction.
Strategy: The average speed is the distance divided by elapsed time.
Solution: Divide the distance by the time:
s=
distance 100.0 m
1 mi 3600 s
=
= 1.830 m/s ร
ร
= 4.095 mi/h
time
54.64 s
1609 m 1 h
Insight: The displacement would be zero in this case because the swimmer swims either two lengths of a 50-m pool or
four lengths of a 25-m pool, returning to the starting point each time. However, the average speed depends upon
distance traveled, not displacement.
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portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
2โ1
James S. Walker, Physics, 5th Edition
Chapter 2: One-Dimensional Kinematics
121. Estimate how fast your hair grows in miles per hour.
Picture the Problem: Your hair grows at a fixed speed.
Strategy: The growth rate is the length gained divided by the time elapsed. Hair grows at a rate of about half an inch a
month, or about 1 cm or 0.01 m per month.
Solution: Divide the length gained by the elapsed time: s =
d 0.010 m 1 mi
1 mo 1 d
=
ร
ร
ร
= 8.5 ร10โ9 mi/h
1 mo 1609 m 30.5 d 24 h
t
Insight: Try converting this growth rate to a more appropriate unit such as ยตm/h. (Answer: 14 ยตm/h.) Choosing an
appropriate unit can help you communicate a number more effectively.
122. IP You drive in a straight line at 20.0 m/s for 10.0 minutes, then at 30.0 m/s for another 10.0 minutes. (a) Is your
average speed 25.0 m/s, more than 25.0 m/s, or less than 25.0 m/s? Explain. (b) Verify your answer to part (a) by
calculating the average speed.
Picture the Problem: You travel in a straight line at two different speeds during the specified time interval.
Strategy: Determine the average speed by first calculating the total distance traveled and then dividing it by the total
time elapsed.
Solution: 1. (a) Because the time intervals are the same, you spend equal times at 20 m/s and 30 m/s, and your average
speed will be equal to 25.0 m/s.
2. (b) Divide the total distance
by the time elapsed:
( 20.0 m/s )(10.0 min ร 60 s ) + ( 30.0 m/s )( 600 s )
s ฮt + s ฮt
sav = 1 1 2 2 =
ฮt1 + ฮt2
600 + 600 s
sav = 25.0 m/s
Insight: The average speed is a weighted average according to how much time you spend traveling at each speed.
123. IP You drive in a straight line at 20.0 m/s for 10.0 miles, then at 30.0 m/s for another 10.0 miles. (a) Is your average
speed 25.0 m/s, more than 25.0 m/s, or less than 25.0 m/s? Explain. (b) Verify your answer to part (a) by calculating the
average speed.
Picture the Problem: You travel in a straight line at two different speeds during the specified time interval.
Strategy: Determine the average speed by first calculating the total distance traveled and then dividing it by the total
time elapsed.
Solution: 1. (a) The distance intervals are the same but the time intervals are different. You will spend more time at
the lower speed than at the higher speed. Because the average speed is a time weighted average, it will be less than
25.0 m/s.
2. (b) Divide the total distance by the time elapsed:
sav =
d1 + d 2
d + d2
20.0 mi
= 1
=
d1 d 2
10.0 mi
10.0 mi ๏ถ
ฮ t1 + ฮ t 2
๏ฆ
+
+
๏ง
๏ท
s1 s 2
20.0
m/s
30.0
m/s ๏ธ
๏จ
sav = 24.0 m/s
Insight: Notice that in this case it is not necessary to convert miles to meters in both the numerator and denominator
because the units cancel out and leave m/s in the numerator.
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portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
2โ2
James S. Walker, Physics, 5th Edition
Chapter 2: One-Dimensional Kinematics
124. CE Predict/Explain Two bows shoot identical arrows with the same launch speed. To accomplish this, the string in
bow 1 must be pulled back farther when shooting its arrow than the string in bow 2. (a) Is the acceleration of the arrow
shot by bow 1 greater than, less than, or equal to the acceleration of the arrow shot by bow 2? (b) Choose the best
explanation from among the following:
I. The arrow in bow 2 accelerates for a greater time.
II. Both arrows start from rest.
III. The arrow in bow 1 accelerates for a greater time.
Picture the Problem: Two arrows are launched by two different bows.
Strategy: Use the definitions of average speed and acceleration to compare the motions of the two arrows.
Solution: 1. (a) We can reason that because both arrows undergo uniform acceleration between the same initial and
final velocities, both arrows must have the same average speed. If they have the same average speed, then arrow 1,
which must travel a longer distance, will be accelerated for a longer period of time. We conclude that the acceleration of
the arrow shot by bow 1 is less than the acceleration of the arrow shot by bow 2.
2. (b) As discussed above, the best explanation is III. The arrow in bow 1 accelerates over a greater time. Statement I is
false and statement II is true but is not a complete explanation.
Insight: We could also set v 0 = 0 in the equation, v 2 = v 02 + 2 a ฮ x and solve for a: a = v 2 2 ฮ x From this expression
we can see that for the same final velocity v, the arrow that is accelerated over the greater distance ฮx will have the
smaller acceleration.
125. IP In the previous problem, (a) does the distance needed to stop increase by a factor of two or a factor of four? Explain.
Verify your answer to part (a) by calculating the stopping distances for initial speeds of (b) 16 m/s and (c) 32 m/s.
Picture the Problem: The car travels in a straight line in the positive direction while accelerating in the negative
direction (slowing down).
Strategy: Use the average velocity and the time elapsed to determine the distance traveled for the specified change in
velocity.
Solution: 1. (a) Because the distance traveled is proportional to the square of the time (Equation 2-11), or alternatively,
because both the time elapsed and the average velocity change by a factor of two, the stopping distance will increase by
a factor of four when you double your driving speed.
2. (b) Evaluate Equation 2-10 directly:
ฮ x = 12 ( v0 + v ) t = 12 (16 + 0 m/s )( 3.8 ) = 30 m = 0.030 km
3. (c) Evaluate Equation 2-10 directly:
ฮ x = 12 ( v0 + v ) t = 12 ( 32 + 0 m/s )( 7.6 ) = 120 m = 0.12 km
Insight: Doubling your speed will quadruple the stopping distance for a constant acceleration. We will learn in chapter
7 that this can be explained in terms of energy; that is, doubling your speed quadruples your kinetic energy.
126. Suppose the car in Problem 44 comes to rest in 35 m. How much time does this take?
Picture the Problem: The car travels in a straight line toward the west while accelerating in the easterly direction
(slowing down).
Strategy: The average velocity is simply half the sum of the initial and final velocities because the acceleration is
uniform. Use the average velocity together with Equation 2-10 to find the time.
Solution: Solve Equation 2-10 for time:
t= 1
ฮx
2 ( v0 + v )
= 1
35 m
2 (12 + 0 m/s )
= 5.8 s
Insight: The distance traveled is always the average velocity multiplied by the time. This stems from the definition of
average velocity.
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portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
2โ3
James S. Walker, Physics, 5th Edition
Chapter 2: One-Dimensional Kinematics
127. Air Bags Air bags are designed to deploy in 10 ms. Estimate the acceleration of the front surface of the bag as it
expands. Express your answer in terms of the acceleration of gravity g.
Picture the Problem: An air bag expands outward with constant positive acceleration.
Strategy: Assume the air bag has a thickness of 1 ft or about 0.3 m. It must expand that distance within the given time
of 10 ms. Employ the relationship between acceleration, displacement, and time (Equation 2-11) to find the
acceleration.
Solution: Solve Equation 2-11 for a:
a=
2( 0.3 m)
2ฮx
1g
=
= 6000 m/s2 ร
๏ป 600g
2
2
t
9.81 m/s2
(10 ms ร 0.001 s/ms)
Insight: The very large acceleration of an expanding airbag can cause severe injury to a small child whose head is too
close to the bag when it deploys. Children are safest in the back seat!
128. IP Coasting due west on your bicycle at 8.4 m/s, you encounter a sandy patch of road 7.2 m across. When you leave the
sandy patch your speed has been reduced by 2.0 m/s to 6.4 m/s. (a) Assuming the sand causes a constant acceleration,
what was the bicycleโs acceleration in the sandy patch? Give both magnitude and direction. (b) How long did it take to
cross the sandy patch? (c) Suppose you enter the sandy patch with a speed of only 5.4 m/s. Is your final speed in this
case 3.4 m/s, more than 3.4 m/s, or less than 3.4 m/s? Explain.
Picture the Problem: A bicycle travels in a straight line, slowing down at a uniform rate as it crosses the sandy patch.
Strategy: Use the time-free relationship between displacement, velocity, and acceleration (Equation 2-12) to find the
acceleration. The time can then be determined from the average velocity and the distance across the sandy patch.
v 2 โ v02 ( 6.4 m/s ) โ ( 8.4 m/s )
=
= โ2.1 m/s 2
a=
2ฮ x
2 ( 7.2 m )
2
Solution: 1. (a) Calculate the acceleration:
2
where the negative sign means 2.1 m/s2 to the east.
t= 1
2. (b) Solve Equation 2-10 for t:
ฮx
2 ( v + v0 )
7.2 m
= 1
2 ( 8.4 + 6.4 m/s )
= 0.97 s
3. (c) Examining v 2 = v 02 + 2 a ฮ x (Equation 2-12) in detail, we note that the acceleration is negative, and that the final
velocity is the square root of the difference between v 02 and 2 a ฮ x . Because 2 a ฮ x is constant because the sandy
patch doesnโt change, it now represents a larger fraction of the smaller v 02 , and the final velocity v will be more than
2.0 m/s different than v 0 . We therefore expect a final speed of less than 3.4 m/s.
Insight: In fact, if you try to calculate v in part (c) with Equation 2-12 you end up with the square root of a negative
02 โ v02 โ ( 5.4 m/s )
=
= 6.9 m , less than the 7.2 m
number, because the bicycle will come to rest in a distance ฮx =
2a
2 ( โ2.1 m/s2 )
2
length of the sandy patch.
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portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
2โ4
James S. Walker, Physics, 5th Edition
Chapter 2: One-Dimensional Kinematics
129. In a physics lab, students measure the time it takes a small cart to slide a distance of 1.00 m on a smooth track inclined
at an angle above the horizontal. Their results are given in the following table.
ฮธ
time, s
10.0ยฐ
1.08
20.0ยฐ
0.770
30.0ยฐ
0.640
(a) Find the magnitude of the cartโs acceleration for each angle.
(b) Show that your results for part (a) are in close agreement with the formula, a = g sin ฮธ. (We will derive this formula
in Chapter 5.)
Picture the Problem: The cart slides down the inclined track,
each time traveling a distance of 1.00 m along the track.
Strategy: The distance traveled by the cart is given by the
constant-acceleration equation of motion for position as a function
of time (Equation 2-11), where x0 = v0 = 0 . The magnitude of
the acceleration can thus be determined from the given distance
traveled and the time elapsed in each case. We can then make the comparison with a = g sin ฮธ .
Solution: 1. Find the acceleration from
Equation 2-11:
x = 0 + 0 + 12 at 2 ๏ a =
2. Now find the values for ฮธ = 10.0ยฐ:
a=
3. Now find the values for ฮธ = 20.0ยฐ:
a=
4. Now find the values for ฮธ = 30.0ยฐ:
a=
2.00 m
(1.08 s )
2
a = g sin ฮธ
a = ( 9.81 m/s2 ) sin10.0๏ฏ = 1.70 m/s2
2
= 3.37 m/s 2
a = ( 9.81 m/s2 ) sin 20.0๏ฏ = 3.35 m/s2
2
= 4.88 m/s 2
a = ( 9.81 m/s2 ) sin10.0๏ฏ = 4.91 m/s2
2.00 m
( 0.640 s )
ฮธ
= 1.71 m/s 2
2.00 m
( 0.770 s )
2x
t2
1.00 m
Insight: We see very good agreement between the formula a = g sin ฮธ and the measured acceleration. The
experimental accuracy gets more and more difficult to control as the angle gets bigger because the elapsed times
become very small and more difficult to measure accurately. For this reason Galileoโs experimental approach (rolling
balls down an incline with a small angle) gave him an opportunity to make accurate observations about free fall without
fancy electronic equipment.
130. Legend has it that Isaac Newton was hit on the head by a falling apple, thus triggering his thoughts on gravity.
Assuming the story to be true, estimate the speed of the apple when it struck Newton.
Picture the Problem: An apple falls straight downward under the influence of gravity.
Strategy: The distance of the fall is estimated to be about 3.0 m (about 10 ft). Then use the time-free equation of
motion (Equation 2-12) to estimate the speed of the apple.
Solution: 1. Solve Equation 2-12 for v,
assuming the apple drops from rest ( v 0 = 0 ):
v = 0 + 2aฮx
2. Let a = g and calculate v:
v = 2 ( 9.81 m/s 2 ) ( 3.0 m ) = 7.7 m/s = 17 mi/h
Insight: Newton supposedly then reasoned that the same force that made the apple fall also keeps the Moon in orbit
around the Earth, leading to his universal law of gravity (Chapter 12). One lesson we might learn here isโwear a
helmet when sitting under an apple tree!
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portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
2โ5
James S. Walker, Physics, 5th Edition
Chapter 2: One-Dimensional Kinematics
131. Jordanโs Jump Michael Jordanโs vertical leap is reported to be 48 inches. What is his takeoff speed? Give your answer
in meters per second.
Picture the Problem: Michael Jordan jumps vertically, the acceleration of gravity slowing him down and bringing him
momentarily to rest at the peak of his flight.
Strategy: Because the height of the leap is known, use the time-free equation of motion (Equation 2-12) to find the
takeoff speed.
Solution: Solve Eq. 2-12 for v 0 :
v0 = v 2 โ 2 g ฮx = 02 โ 2 ( โ9.81 m/s 2 ) ( 48 in ร 0.0254 m/in ) = 4.9 m/s
Insight: That speed is about half of what champion sprinters achieve in the horizontal direction, but is very good among
athletes for a vertical leap. High jumpers can jump even higher, but use the running start to their advantage.
132. Bill steps off a 3.0-m-high diving board and drops to the water below. At the same time, Ted jumps upward with a
speed of 4.2 m/s from a 1.0-m-high diving board. Choosing the origin to be at the waterโs surface, and upward to be the
positive x direction, write x-versus-t equations of motion for both Bill and Ted.
Picture the Problem: Two divers move vertically under the influence of gravity.
Strategy: In both cases we wish to write the equation of motion for position as a function of time and acceleration
(Equation 2-11). In Billโs case, the initial height x0 = 3.0 m , but the initial velocity is zero because he steps off the
diving board. In Tedโs case the initial height x0 = 1.0 m and the initial velocity is +4.2 m/s. In both cases the
acceleration is โ9.81 m/s2.
Solution: 1. Equation 2-11 for Bill:
x = x0 + v0 t + 12 at 2 = 3.0 m + 0 + 12 ( โ9.81 m/s 2 ) t 2
x = ( 3.0 m ) โ ( 4.9 m/s 2 ) t 2
2. Equation 2-11 for Ted:
x = x0 + v0 t + 12 at 2 = 1.0 m + ( 4.2 m/s ) t + 12 ( โ9.81 m/s 2 ) t 2
x = (1.0 m ) + ( 4.2 m/s ) t โ ( 4.9 m/s 2 ) t 2
Insight: The different initial velocities result in significantly different trajectories for Bill and Ted.
133. Repeat the previous problem, this time with the origin 3.0 m above the water, and with downward as the positive
x direction.
Picture the Problem: Two divers move vertically under the influence of gravity.
Strategy: In both cases we wish to write the equation of motion for position as a function of time and acceleration
(Equation 2-11). Here weโll take the origin to be at the level of Billโs board above the water, Tedโs diving board to be at
+2.0 m, and the water surface at +3.0 m. Downward is the positive direction so that the acceleration is 9.81 m/s2. In
Billโs case, the initial height x0 = 0.0 m and his initial velocity is zero because he steps off the diving board. In Tedโs
case the initial height is x0 = +2.0 m and the initial velocity is โ 4.2 m/s (upward).
Solution: 1. Equation 2-11 for Bill:
x = x0 + v0 t + 12 at 2 = 0.0 m + 0 + 12 ( 9.81 m/s 2 ) t 2
x = ( 4.9 m/s 2 ) t 2
2. Equation 2-11 for Ted:
x = x0 + v0 t + 12 at 2 = 2.0 m + ( โ4.2 m/s ) t + 12 ( 9.81 m/s 2 ) t 2
x = ( 2.0 m ) + ( โ 4.2 m/s ) t + ( 4.9 m/s 2 ) t 2
Insight: The different initial velocities result in significantly different trajectories for Bill and Ted.
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portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
2โ6
James S. Walker, Physics, 5th Edition
Chapter 2: One-Dimensional Kinematics
134. IP Standing side by side, you and a friend step off a bridge at different times and fall for 1.6 s to the water below. Your
friend goes first, and you follow after she has dropped a distance of 2.0 m. (a) When your friend hits the water, is the
separation between the two of you 2.0 m, less than 2.0 m, or more than 2.0 m? (b) Verify your answer to part (a) with a
calculation.
Picture the Problem: You and your friend both accelerate
from rest straight downward, but at different times. You step
off the bridge when your friend has fallen 2.0 m, and your
friend hits the water while you are still in the air.
you
jump
bridge
2.0 m
Strategy: First find the time it takes for your friend to fall
2.0 m using the equation of motion for position as a function
of time and acceleration (Equation 2-11). Subtract that time
from 1.6 s to find the time elapsed between when you jump
and when your friend hits the water. Use Equation 2-11 and
the times found above to find the positions of you and your
friend at the time your friend lands. Then determine the
separation between the known positions.
friend
lands
t = 1.6 s
S?
water
Solution: 1. (a) Because your friend has a greater average speed than you do during the time between when you jump
and your friend lands, the separation between the two of you will increase to a value more than 2.0 m.
2 ( 2.0 m )
2ฮ x
=
= 0.64 s
g
9.81 m/s 2
2. (b) Find the time it takes to fall 2.0 m from
Equation 2-11 with v 0 = 0 :
t=
3. Find the distance your friend fell in 1.6 s:
xfriend = 12 gt 2 = 12 ( 9.81 m/s 2 ) (1.6 s ) = 13 m
4. Find the distance you fell in the shorter time:
xyou = 12 g ( t โ t2.0 m ) = 12 ( 9.81 m/s 2 ) (1.6 โ 0.64 s ) = 4.5 m
5. Find the difference in your positions:
S = xfriend โ xyou = 13 โ 4.5 m = 8 m
2
2
2
Insight: Because of her head start, your friend will always have a higher average velocity than you, and the separation
between you and her will continue to increase the longer you both fall.
135. In a well-known Jules Verne novel, Phileas Fogg travels around the world in 80 days. What was Mr. Foggโs
approximate average speed during his adventure?
Picture the Problem: Phileas Fogg travels in a straight line all the way around the world.
Strategy: The average speed is the distance divided by elapsed time. We will estimate that Mr. Fogg travels a distance
equal to the equatorial circumference of the Earth. This is an approximation, because his path was most likely much
more complicated than that, but we were asked only for the approximate speed.
Solution: Find the circumference of the Earth:
d = 2ฯ r = 2ฯ ( 6370 ร 103 m ) = 4.0 ร 107 m
Divide the distance by the time:
s=
distance
4.0 ร 107 m
=
= 5.8 m/s
time
80 d ร 24 h/d ร 3600 s/h
Insight: This speed corresponds to about 13 mi/h and is faster than humans can walk. Giving time for sleeping, eating,
and other delays, Mr. Fogg needs a relatively fast means of travel.
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portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
2โ7
James S. Walker, Physics, 5th Edition
Chapter 2: One-Dimensional Kinematics
136. You jump from the top of a boulder to the ground 1.5 m below. Estimate your deceleration on landing.
Picture the Problem: You jump off a boulder, accelerate from rest straight downward and land, bending your knees so
that your center of mass comes to rest over a short vertical distance.
Strategy: Employ the relationship between acceleration, displacement, and velocity (Equation 2-12) to find your final
velocity just before landing. Then estimate the distance your center of mass will move after your feet contact the
ground, and use that distance to estimate your deceleration rate.
Solution: 1. Solve Equation 2-12 for velocity v:
v = v02 + 2aฮx = 02 + 2 ( 9.81 m/s2 ) (1.5 m ) = 5.4 m/s
2. Estimate your center of mass moves downward
about 0.5 m after your feet contact the ground and
you bend your knees into a crouching position.
Solve Equation 2-12 for acceleration:
a=
v 2 โ v02 0 โ ( 5.4 m/s )
=
= โ29 m/s 2 = โ3.0 g
2ฮy
2 ( 0.50 m )
2
2
Insight: When a gymnast lands from an even higher altitude, she might try to bend her knees even less in order to
impress the judges. If she lands from an altitude of 3.0 m and bends her knees so her center of mass moves only 0.2 m,
her acceleration is โ15g!
137. CE At the edge of a roof you drop ball A from rest, and then throw ball B downward with an initial velocity of v0 . Is
the increase in speed just before the balls land more for ball A, more for ball B, or the same for each ball?
Picture the Problem: Two balls are released from the edge of a roof. Ball A is dropped from rest but ball B is thrown
downward with an initial velocity v0 .
Strategy: Use the definition of acceleration to answer the conceptual question, keeping in mind the average speed of
ball B is greater than the average speed of ball A.
Solution: The two balls fall the same distance but ball B has the greater average speed and falls for a shorter length of
time. Because each ball accelerates at the same rate of 9.81 m/s2, ball A accelerates for a longer time and the increase in
speed is more for ball A than it is for ball B.
Insight: If ball B were fired downward at an extremely high speed, it would reach the ground within a very short
interval of time and its speed would hardly change at all.
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portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
2โ8
James S. Walker, Physics, 5th Edition
Chapter 2: One-Dimensional Kinematics
138. IP A youngster bounces straight up and down on a trampoline. Suppose she doubles her initial speed from 2.0 m/s to
4.0 m/s. (a) By what factor does her time in the air increase? (b) By what factor does her maximum height increase?
(c) Verify your answers to parts (a) and (b) with an explicit calculation.
Picture the Problem: A youngster bounces straight up and down on a trampoline. The child rises straight upward,
slows down, and momentarily comes to rest before falling straight downward again.
Strategy: Find the time of flight by exploiting the symmetry of the situation. If it takes time t for gravity to slow the
child down from her initial speed v0 to zero, it will take the same amount of time to accelerate her back to the same
speed. She therefore lands at the same speed v0 with which she took off. Use this fact together with Equation 2-7 to
find the time of flight. The maximum height she achieves is related to the square of v0, as indicated by Equation 2-12.
Solution: 1. (a) Because the time of flight depends linearly upon the initial velocity, doubling v0 will increase her time
of flight by a factor of 2.
2. (b) Because the time of flight depends upon the square of the initial velocity, doubling v0 will increase her maximum
altitude by a factor of 4.
3. (c) The time of flight for v0 = 2.0 m/s , using Eq. 2-7:
t=
v โ v0 ( โ v0 ) โ v0 2v0 2 ( 2.0 m/s )
=
=
=
= 0.41 s
โg
โg
g
9.81 m/s 2
4. The time of flight for v0 = 4.0 m/s :
t=
2 v0 2 ( 4.0 m/s )
=
= 0.82 s
g
9.81 m/s 2
5. The maximum height for v0 = 2.0 m/s , using Eq. 2-12:
ฮx =
6. The maximum height for v0 = 4.0 m/s :
v2
( 4.0 m/s )
ฮx = 0 =
= 0.82 m
2 g 2 ( 9.81 m/s 2 )
v 2 โ v02 02 โ v02 v02
( 2.0 m/s )
=
=
=
= 0.20 m
โ2 g
โ2 g
2 g 2 ( 9.81 m/s2 )
2
2
Insight: The reason the answer in step 6 is not exactly four times larger than the answer in step 5 is due to the rounding
required by the fact that there are only two significant digits. If you recalculate using 2.00 m/s and 4.00 m/s, the
answers are 0.204 and 0.816 m, respectively.
139. IP A popular entertainment at some carnivals is the blanket toss (see photo, p. 39). (a) If a person is thrown to a
maximum height of 28.0 ft above the blanket, how long does she spend in the air? (b) Is the amount of time the
person is above a height of 14.0 ft more than, less than, or equal to the amount of time the person is below a height
of 14.0 ft? Explain. (c) Verify your answer to part (b) with a calculation.
Picture the Problem: The person is thrown straight upward, slows down, and momentarily comes to rest before falling
straight downward again.
Strategy: Find the time of flight by exploiting the symmetry of the situation. If it takes time t for gravity to slow the
person down from her initial speed v0 to zero, it will take the same amount of time to accelerate her back to the same
speed. It therefore takes the same amount of time for her to rise to the peak of her flight than it does for her to return to
the blanket. Use this fact together with Equation 2-11 with v0 = 0 (corresponding to the second half of her flight, from
the peak back down to the blanket) to find the time of flight. The time above and below 14.0 ft can be found using the
same equation.
Solution: 1. (a) The time of flight
can be found from Equation 2-11:
t = 2 ร tdown = 2 ร
2 ( 28.0 ft ร 0.305 m/ft )
2ฮ x
=2
= 2.64 s
g
9.81 m/s 2
2. (b) The personโs average speed is less during the upper half of her trajectory, so the time she spends in that portion of
her flight is more than the time she spends in the lower half of her flight.
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portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
2โ9
James S. Walker, Physics, 5th Edition
Chapter 2: One-Dimensional Kinematics
2 (14.0 ft ร 0.305 m/ft )
2ฮ x
=2
= 1.87 s
g
9.81 m/s 2
3. (c) The time she spends above 14.0 ft
is the same time of her flight if her
maximum height were 14.0 ft:
tabove = 2 ร
4. The time spent below 14.0 ft is the
remaining portion of the total time of flight:
tbelow = ttotal โ tabove = 2.64 โ 1.87 s = 0.77 s
Insight: The symmetry of the motion of a freely falling object can often be a useful tool for solving problems quickly.
140. Referring to Conceptual Checkpoint 2โ5, find the separation between the rocks at (a) t = 1.0 s, (b) t = 2.0 s, and
(c) t = 3.0 s, where time is measured from the instant the second rock is dropped. (d) Verify that the separation
increases linearly with time.
Picture the Problem: The two rocks fall straight downward along a similar path except at different
times.
Strategy: First find the time elapsed between the release of the two rocks by finding the time
required for the first rock to fall 4.00 m, using the equation of motion for position as a function of
time and acceleration (Equation 2-11). The positions as a function of time for each rock can then be
compared to find a separation distance as a function of time.
Solution: 1. (a) Find the time required
for rock A to fall 4.00 m:
2. Let t represent the time elapsed from
the instant rock B is dropped. The
position of rock A (Equation 2-11) is
thus:
t4 =
2ฮ x
=
g
2 ( 4.00 m )
9.81 m/s 2
= 0.903 s
x A = 0 + 12 g ( t + t4 ) = 12 gt 2 + g t t4 + 12 gt42
2
4. The position of rock B (Equation 2-11)
is:
xB = 0 + 12 gt 2 = 12 gt 2
5. Find the separation between the rocks:
ฮ x = x A โ xB = ( 12 gt 2 + g t t4 + 12 gt42 ) โ 12 gt 2
ฮ x = g t t4 + 12 gt42 = ( 9.81 m/s 2 ) t ( 0.903 s ) + 12 ( 9.81 m/s 2 ) ( 0.903 s )
2
ฮ x = ( 8.86 m/s ) t + 4.00 m
6. Find ฮx for t = 1.0 s:
ฮ x = ( 8.86 m/s )(1.0 s ) + 4.00 m = 12.9 m
7. (b) Find ฮx for t = 2.0 s:
ฮ x = ( 8.86 m/s )( 2.0 s ) + 4.00 m = 22 m
8. (c) Find ฮx for t = 1.0 s:
ฮ x = ( 8.86 m/s )( 3.0 s ) + 4.00 m = 31 m
9. (d) The linear dependence of ฮ x upon t can be verified by examining the equation derived in step 5.
Insight: The only way for rock B to catch up to rock A would be for rock B to be thrown downward with a large initial
(
)
speed. In that case the separation becomes ฮx = 8.86 m/s โ vB,0 t + 4.00 m, which decreases to zero as long as vB,0 is
greater than 8.86 m/s.
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portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
2 โ 10
James S. Walker, Physics, 5th Edition
Chapter 2: One-Dimensional Kinematics
141. In the previous problem, what is the minimum initial speed of the camera if it is to just reach the passenger?
(Hint: When the camera is thrown with its minimum speed, its speed on reaching the passenger is the same
as the speed of the passenger.)
Picture the Problem: The trajectories of the balloon and
camera are shown at right. The balloon rises at a steady rate
while the cameraโs speed is continually slowing down under
the influence of gravity. The camera is caught when the two
trajectories meet.
Strategy: The camera meets the balloon when the positions are
equal, so that is our starting point. For the case when the
camera just barely meets the balloon, the velocity of the camera
must match the velocity of the balloon (2.0 m/s). We use this
fact to find the time the two must meet, and substitute that into
the position equation. We can then solve for the initial velocity
of the camera.
Solution: 1. Write Equation 2-10 for the balloon:
xb = xb,0 + vb t
2. Write Equation 2-12 for the camera:
xc =
3. Set xb = xc and solve for vc ,0 :
xb ,0 + vb t =
4. As indicated above, the camera will be caught not
only when itโs at the same position as the balloon, but
when its velocity is the same as well, so set v c = v b :
v c2, 0 = v b2 + 2 gx b , 0 + 2 gv b t
v c2 โ v c2,0
โ2 g
vc2 โ vc2,0
โ2 g
๏ vc2,0 = vc2 + 2 g ( xb ,0 + vb t )
5. The two will meet at a time when their velocities are
equal. Write Equation 2-7 for the camera and set its
final velocity equal to the balloonโs velocity, and find
the time.
v โv
t = c,0 b
g
6. Substitute the time into the equation in step 4:
vc2,0 = vb2 + 2 gxb ,0 + 2vb ( vc ,0 โ vb )
vc = vc,0 โ gt = vb
vc2,0 โ 2vb vc ,0 + vb2 โ 2 gxb ,0 = 0
vc2,0 โ 2 ( 2.0 m/s ) vc ,0 + ( 2.0 m/s ) โ 2 ( 9.81 m/s 2 ) ( 2.5 m ) = 0
2
vc2,0 โ 4.0vc ,0 โ 45 m 2 /s 2 = 0
7. You can get the roots using the quadratic formula,
but you might recognize the simple factors here. Only
the positive root corresponds to the camera going
upward:
( vc + 5)( vc โ 9 ) = 0
vc = โ5.0, 9.0 m/s
Insight: This is a complicated problem that always ends with a quadratic solution. It required the kind of strategy that
must usually be mapped out after trying a few things; donโt feel bad if you didnโt intuitively choose this strategy. There
are other strategies that work, but they are equally complicated.
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portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
2 โ 11
James S. Walker, Physics, 5th Edition
Chapter 2: One-Dimensional Kinematics
142. Old Faithful Watching Old Faithful erupt, you notice that it takes a time t for water to emerge from the base of the
geyser and reach its maximum height. (a) What is the height of the geyser, and (b) what is the initial speed of the water?
Evaluate your expressions for (c) the height and (d) the initial speed for a measured time of 1.65 s.
Picture the Problem: The water shoots straight upward, slows down, and momentarily comes to rest before falling
straight downward again.
Strategy: Find the height of the geyser by exploiting the symmetry of the situation. If it takes time t for gravity to slow
the water down from its initial speed v0 to zero, it will take the same amount of time to accelerate it back to the same
speed. The height of the geyser is therefore determined by the distance the water will fall from rest in time t (Equation
2-11). Gravity will slow the water down from its initial velocity to zero in time t at a known rate ( โ 9.81 m /s 2 ), so that
fact can be used to find the initial velocity (Equation 2-7).
Solution: 1. (a) Solve Equation 2-11 for x0, setting x = 0
and v 0 = 0 for the case when the water falls from rest in time t:
0 = x0 + 0 โ 12 g t 2
2. (b) Use Equation 2-7 to find the initial velocity if the final
velocity is zero (upward portion of the flight):
v = v0 โ g t = 0
3. (c) Substitute t = 1.65 s into the equation from step 1:
xmax = 12 ( 9.81 m/s 2 ) (1.65 s ) = 13.4 m
4. (d) Substitute t = 1.65 s into the equation from step 2:
v0 = ( 9.81 m/s 2 ) (1.65 s ) = 16.2 m/s
xmax = x0 = 12 g t 2
v0 = g t
2
Insight: If you round off g = 10 m/s2, you can impress your friends by memorizing these simple formulae and doing the
quick calculations in your head!
143. IP A ball is thrown upward with an initial speed v0 . When it reaches the top of its flight, at a height h, a second ball is
thrown upward with the same initial velocity. (a) Sketch an x-versus-t plot for each ball. (b) From your graph, decide
whether the balls cross paths at h/2, above h/2, or below h/2. (c) Find the height where the paths cross.
Picture the Problem: The trajectories of the two balls are shown at
right. Remember that in each case the balls are traveling straight up and
straight down; the graphs look parabolic because time is the x axis. Ball
B is tossed upward at the instant ball A reaches the peak of its flight. Ball
A has begun its descent when it is passed by ball B, which is still on its
way up toward its peak.
Strategy: The positions are equal to each other when the balls cross
paths. The launch times are offset by the time it takes the ball to reach
the peak of its flight. That time is given by the time it takes gravity to
slow the ball from v0 down to zero (Equation 2-7). The time the balls
cross is directly between the time ball B is launched and ball A lands.
Once we have the time figured out we can find the position of ball A in
terms of its maximum height h.
Solution: 1. The plot of x-versus-t for the two balls is shown above.
2. Judging from the plot the balls will cross paths above h / 2.
t=
3. Find the time it takes ball A to reach its peak:
v โ v0 0 โ v0 v0
=
=
โg
โg
g
4. Because ball B is launched at time v0 g and ball A lands at time 2v0 g , the two balls will cross at a time midway
between these, or at time tcross = 3v0 2 g .
2
5. Find the position of ball A at time tcross using Eq. 2-11:
x A = v0 tcross โ gt
1
2
2
cross
๏ฆ 3v ๏ถ
๏ฆ 3v ๏ถ
3v 2
= v0 ๏ง 0 ๏ท โ 12 g ๏ง 0 ๏ท = 0
8g
๏จ 2g ๏ธ
๏จ 2g ๏ธ
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2 โ 12
James S. Walker, Physics, 5th Edition
Chapter 2: One-Dimensional Kinematics
6. Find the maximum height h using Equation 2-12:
0 2 = v02 โ 2 gh ๏ h =
7. Now write x A in terms of h:
x A 3v02 8 g 3
= 2
=
h
v0 2 g
4
v02
2g
๏ x A = 34 h
Insight: The balls do not cross right at h / 2 because they spend more time above h / 2 than they do below, because their
average speeds are smaller during the top half of their flight.
144. Weights are tied to each end of a 20.0-cm string. You hold one weight in your hand and let the other hang vertically a
height h above the floor. When you release the weight in your hand, the two weights strike the ground one after the
other with audible thuds. Find the value of h for which the time between release and the first thud is equal to the time
between the first thud and the second thud.
Picture the Problem: The two weights fall straight downward from rest along a similar path except
at different times.
Strategy: The problem requires that the time to fall a distance h from rest (the time between release
and the first thud) is the time to fall a distance h + 20 cm (second thud) minus the time to fall a
distance h (first thud). We can set these times equal to each other, use Equation 2-11 to write the
times in terms of heights, and then solve for h.
Solution: 1. Set the time
intervals equal to each other:
th = th + 20 โ th
2. Now use Equation 2-11 to write
the times in terms of the heights:
2
3. Square both sides and multiply by g / 2:
4h = h + 20.0 cm
2h
=
g
20 cm
h
๏ 2th = th + 20
2 ( h + 20.0 cm )
g
20.0
cm = 6.67 cm
3
Insight: The tension in the string will be zero during the descent because each ball accelerates at the same rate.
Therefore the string will have no effect upon the motion of the balls.
h=
145. A stalactite on the roof of a cave drips water at a steady rate to a pool 4.0 m below. As one drop of water hits the pool, a
second drop is in the air, and a third is just detaching from the stalactite. (a) What are the position and velocity of the
second drop when the first drop hits the pool? (b) How many drops per minute fall into the pool?
Picture the Problem: The three drops are positioned as depicted at right. They all fall straight
downward from an initial height of 4.0 m.
Strategy: The time interval between drops is half the time it takes a drop to fall the entire 4.0 m. Use
this fact to find the position and velocity of drop 2 when drop 1 hits the pool (equations 2-11 and 2-7).
Then the time interval between drops can be used to find the number of drops per minute.
Solution: 1. (a) Find the time interval between
drops, using Equation 2-11 to find the fall time:
2. Now use Equation 2-11 to find the position
of drop 2:
ฮ t = 12 tfall =
3
1 2 x 1 2 ( 4.0 m )
=
= 0.45 s
2 g
2 9.81 m/s 2
x2 = 0 + g ( ฮ t ) =
1
2
stalactite
2
1
2
( 9.81 m/s ) ( 0.45 s )
2
2
x2 = 0.99 m below the stalactite or
2
4.0 m
x2
1
4.0 โ 0.99 m = 3.0 m above the pool
3. Use Equation 2-7 to find the speed of drop 2:
4. (b) Find the drop rate from the time interval:
v = 0 + g ฮ t = ( 9.81 m/s 2 ) ( 0.45 s ) = 4.4 m/s
D=
1 drop 60 s
ร
= 130 drops/min
0.45 s 1 min
Insight: Note that it takes half the drop time to fall the first quarter of the drop distance, and half the time to fall the
final three quarters of the distance.
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portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
2 โ 13
James S. Walker, Physics, 5th Edition
Chapter 2: One-Dimensional Kinematics
146. Suppose the first rock in Conceptual Checkpoint 2โ5 drops through a height h before the second rock is released from
rest. Show that the separation between the rocks, S, is given by the following expression:
S = h + ( 2 gh )t
In this result, the time t is measured from the time the second rock is dropped.
Picture the Problem: The two rocks fall straight downward along a similar path except at different
times.
Strategy: First find the time elapsed between the release of the two rocks by finding the time
required for the first rock to fall a distance h, using the equation of motion for position as a function
of time and acceleration (Equation 2-11). The positions as a function of time for each rock can then
be compared to find a separation distance as a function of time.
Solution: 1. (a) Find the time required for rock A to
fall a distance h:
th =
2ฮx
=
g
2h
g
2. Let t represent the time elapsed from the instant
rock B is dropped. The position of rock A (equation
2-11) is thus:
xA = 0 + 12 g ( t + th ) = 12 gt 2 + g t th + 12 gth2
3. The position of rock B (Equation 2-11) is:
x B = 0 + 12 gt 2 = 12 gt 2
4. Find the separation between the rocks:
S = x A โ xB = ( 12 gt 2 + g t th + 12 gth2 ) โ 12 gt 2
2
S = g t th + 12 gth2 = gt
2h 1 2h
+ g
g 2 g
S = t 2g h + h = h +
( 2 gh ) t
Insight: The separation between the two rocks increases linearly with time t.
147. An arrow is fired with a speed of 20.0 m/s at a block of Styrofoam resting on a smooth surface. The arrow penetrates a
certain distance into the block before coming to rest relative to it. During this process the arrowโs deceleration has a
magnitude of 1550 m/s2 and the blockโs acceleration has a magnitude of 450 m/s2. (a) How long does it take for the
arrow to stop moving with respect to the block? (b) What is the common speed of the arrow and block when this
happens? (c) How far into the block does the arrow penetrate?
Picture the Problem: An arrow travels horizontally at 20.0 m/s and impacts the Styrofoam. It continues to travel in the
positive direction, but more slowly due to its collision with the Styrofoam. The arrow and the Styrofoam then move
together at the same speed in the positive direction.
Strategy: Find the final velocity of the block in terms of the collision time ฮ t by using Equation 2-7. Because this is
also the final velocity of the arrow, the collision time ฮ t can be determined by using the known accelerations and the
initial velocity of the arrow. The final velocity and penetration depth traveled can then be found from applying
equations 2-7 and 2-11.
va = vb
Solution: 1. (a) Set the final velocities of the
arrow and the block equal to each other and
apply Equation 2-7 to find ฮ t :
va ,0 + aa ฮt = 0 + ab ฮt
ฮt =
โ va ,0
aa โ ab
=
va ,0
ab โ aa
=
20.0 m/s
450 โ ( โ1550 ) m/s 2
ฮt = 0.0100 s = 10.0 ms
2. (b) Now apply Equation 2-7 to find v b :
vb = ab ฮ t = ( 450 m/s 2 ) ( 0.0100 s ) = 4.50 m/s
Copyright ยฉ 2017 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No
portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
2 โ 14
Physics 5th Edition Walker Solutions Manual
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Chapterhttp://testbanklive.com/download/physics-5th-edition-walker-solutions-manual/
2: One-Dimensional Kinematics
James S. Walker, Physics, 5th Edition
3. (c) The penetration distance is a bit tricky
because both the arrow and the block move
while they are colliding. The penetration
distance is the difference between how far the
arrow moves and how far the block moves
during the collision time interval.
d = ฮ xarrow โ ฮ xblock
= ( va ,0 ฮ t + 12 aa ฮ t 2 ) โ ( 12 ab ฮt 2 )
๏ฉ( 20.0 m/s )( 0.0100 s ) + 1 ( โ1550 m/s 2 ) ( 0.0100 s )2 ๏น
2
๏บ
=๏ช
2
2
1
๏ช
๏บ
450
m/s
0.0100
s
โ
(
)
(
)
2
๏ซ
๏ป
d = 0.1225 m โ 0.0225 m = 0.100 m = 10.0 cm
Insight: We could also analyze this collision using the concept of momentum conservation (Chapter 9) and work and
energy (Chapter 7).
148. Sitting in a second-story apartment, a physicist notices a ball moving straight upward just outside her window. The ball
is visible for 0.25 s as it moves a distance of 1.05 m from the bottom to the top of the window. (a) How long does it take
before the ball reappears? (b) What is the greatest height of the ball above the top of the window?
Picture the Problem: This exercise considers a generic object traveling in a straight line with constant acceleration.
Strategy: Manipulate the suggested equations with algebra to derive the desired results.
Solution: 1. (a) Begin with Equation 2-12:
v 2 = v 02 + 2 a ( x โ x 0 )
2. Set x = 0 and solve for v:
v = ยฑ v02 โ 2 a x0
v = v0 + at
3. (b) First write Equation 2-7 and
substitute for v. Then solve for t:
ยฑ v02 โ 2a x0 = v0 + at
โv0 ยฑ v02 โ 2a x0
a
=t
0 = x0 + v0 t + 12 at 2
4. (c) Write Equation 2-11 as given and
apply the quadratic formula to solve for t:
t=
t=
โb ยฑ b 2 โ 4ac
2a
=
โv0 ยฑ v02 โ 4 ( 12 a ) ( x0 )
2 ( 12 a )
โv0 ยฑ v02 โ 2a x0
a
Insight: When an object undergoes uniform acceleration its position is a quadratic function of time. The quadratic
formula is therefore an appropriate one to describe the motion of the object.
Copyright ยฉ 2017 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No
portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
2 โ 15
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