Preview Extract

2
Motion in One Dimension
CHAPTER OUTLINE
2.1
Position, Velocity, and Speed of a Particle
2.2
Instantaneous Velocity and Speed
2.3
Analysis Model: Particle Under Constant Velocity
2.4
The Analysis Model Approach to Problem Solving
2.5
Acceleration
2.6
Motion Diagrams
2.7
Analysis Model: Particle Under Constant Acceleration
2.8
Freely Falling Objects
2.9
Kinematic Equations Derived from Calculus
* An asterisk indicates a question or problem new to this edition.
SOLUTIONS TO THINK-PAIR-SHARE AND ACTIVITIES
*TP2.1
Conceptualize The photo in Figure TP2.1 helps you to visualize the situation.
In the first demonstrations, the object, which appears as a small silver-colored
slider at the left end of the red lever in the photograph, rises up and just
touches the bell at the top of the column. We can approximate this situation
by saying that the slider just reaches zero velocity at the position of the bell.
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In the final demonstration, the hitting of the hammer on the apparatus has
caused the bell to shift to the side, so that the slider can fly vertically off the
track when it is projected with a higher initial velocity. It looks like we have
only two pieces of information: the height h of the object in the first situation
and the fact that the initial velocity of the object in the second situation is
twice that in the first. But we have two more pieces of information that can be
gleaned from the description: in the first situation, the velocity of the object
goes to zero just as its vertical position reaches height h; in the second
situation, the initial and final vertical positions are the same.
Categorize The slider is an object in free fall, so we use the particle under
constant acceleration model for its motion.
Analyze For the first situation, write an equation from the model relating
velocity and position, using subscripts 1 to describe this situation. Evaluate
the final values in the equation for the instant at which the object just touches
the bell and momentarily comes to rest:
(
v 2f 1 = vi12 โ 2g y f 1 โ yi1
) โ 0 = v โ 2g ( h โ 0) โ v = 2gh (1)
2
i1
i1
For the second situation, write an equation from the model relating velocity
and time, using subscripts 2. Solve for the time:
v f 2 = vi 2 โ gt โ t =
vi 2 โ v f 2
g
(2)
Now evaluate the time at which the object reaches its highest position at the
top of its motion:
ttop =
vi 2 โ 0 vi 2
=
g
g
(3)
Recognize that the initial velocity in the second situation is twice that of the
first, and combine Equations (1) and (2):
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ttop =
2vi1 2
=
g
( 2gh ) = 2 2h
g
g
Finally, recognize by symmetry that the time at which the object lands back
on the ground is twice that at which it is at the top, and substitute numerical
values:
tground = 2ttop = 4
2 ( 4.50 m )
2h
=4
= 3.83 s
g
9.80 m/ s 2
Finalize Does this time interval for the flight of the object seem reasonable?
Discuss with your partners the following. Is this time just twice the time for
the object to fall back to the lever in the first situation? In the second situation,
does the object rise to a maximum height of just twice that in the first
situation? Does the fact that the object begins its upward motion a few
centimeters above the ground make any significant difference in our
calculation of the final time in the second situation?
Answer: 3.83 s
*TP2.2
Conceptualize Be sure you understand the setup of the problem. One person
drops a rock from rest. The second person waits for a time interval and then
throws the second rock with just the right speed so that it catches up to the
first rock just as the two rocks enter the water.
Categorize The rocks are in free fall, so we use the particle under constant
acceleration model for their motion.
Analyze Let us first look at the fall of the first rock through the vertical
distance h. Using Equation 2.16, rewritten for the vertical direction, we find
y f = yi + v yit + 12 ayt 2
โ
โ h = 0 + 0 โ 12 gt 2
โ t=
2h
g
(1)
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This is the time at which the dropped rock strikes the water. Now consider
the fall of the second rock. The thrower waits for a time interval โt after the
first rock is dropped before throwing his rock downward. Let us denote time
tโ as the time at which the second rock reaches the water, with tโ = 0 at the
instant the second rock is thrown downward. Using Equation 2.16, rewritten
for the vertical direction, we find
y f = yi + v yit โฒ + 12 ay ( t โฒ )
2
h โ 12 g ( t โฒ )
v yi =
tโฒ
2
โ
โ h = 0 โ v yit โฒ โ 12 g ( t โฒ )
โ
2
(2)
In order for there to be a single splash, the time t at which the first rock strikes
the water must be the same as the sum of the waiting time โt and the time tโ
for the second rock to fall:
t = ฮt + t โฒ โ t โฒ = t โ ฮt
(3)
Substitute Equation (3) into Equation (2) and then substitute Equation (1) into
Equation (2):
h โ 12 g ( t โ ฮt )
v yi =
=
t โ ฮt
2
โ 2h
โ
h โ 12 g โ
โ ฮt โ
โ g
โ
2h
โ ฮt
g
2
=
ฮt 2gh โ 12 g ( ฮt )
2
2h
โ ฮt
g
(4)
(a) Substitute numerical values:
(1.00 s) 2 ( 9.80 m/ s ) ( 75.0 m ) โ 12 ( 9.80 m/ s ) (1.00 s)
v =
= 11.5 m/ s
2 ( 75.0 m )
โ 1.00 s
2
2
2
yi
( 9.80 m/ s )
2
(b) One way to solve this problem is to solve Equation (4) for โt. This leads to
a quadratic equation whose appropriate solution is
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ฮt =
v yi + 2gh โ v yi2 + 2gh
(5)
g
Substitute numerical values:
ฮt =
(
)
40.0 m/ s + 2 9.80 m/ s 2 ( 75.0 m ) โ
( 40.0 m/ s) + 2 ( 9.80 m/ s ) ( 75.0 m )
2
2
9.80 m/ s 2
= 2.34 s
Another way to solve the problem is to set up a spreadsheet using Equation
(4) and generate a list of velocities vyi for increasing values of โt. You can then
look in the list for the speed for 40.0 m/s and read the approximate value of
โt. Then the spreadsheet can be modified by looking at just the short range of
โt that gives a speed of about 40.0 m/s and using a smaller increment for โt.
By narrowing in using this technique, you find again that the time interval is
2.34 s.
(c) The second rock must be thrown before the first rock hits the water.
Therefore, the longest time interval is given by Equation (1):
ฮtmax = t =
2h
g
Substitute numerical values:
ฮtmax =
2 ( 75.0 m )
= 3.91 s
9.80 m/ s 2
Finalize If the second person waited to throw the second rock for the time
interval in part (c), it would have to be thrown with infinite speed. (We will find
in Chapter 38 that this is not possible.) If you have set up the spreadsheet for part
(b), you will see the values for the speed vyi become very large as you approach
3.91 s and then become negative for larger values of the time interval.
Answers: (a) 11.5 m/s (b) 2.34 s (c) 3.91 s
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*TP2.3
Conceptualize Think about the falling ruler. When it begins to fall, your eyes
see the motion and send a signal to your brain. Your brain has to process the
information and send a signal to your hand to grab the ruler. The muscles in
your hand then have to respond to this signal and close your fingers onto the
ruler. The time intervals for all of these things to happen are added to equal
the reaction time.
Categorize The ruler falls with an acceleration due to gravity, so we use the
particle under constant acceleration model for its motion.
Analyze There are no numbers given in the problem, so letโs make one up.
Suppose your fingers close on the ruler at the 8.0-inch mark. Then, using
Equation 2.16 in the vertical direction,
y f = yi + v yit + 12 ayt 2
โ
โ h = 0 + 0 โ 12 gt 2
โ t=
2h
g
Substitute the value of h:
t=
2 ( 8.0 in ) โ 0.025 4 m โ
= 0.20 s
9.80 m/ s 2 โโ 1 in โโ
Finalize This value is a typical human reaction time.
Answer: Answers will vary depending on your individual reaction time. A
typical human reaction time is about 0.2 s.
*TP2.4
Conceptualize From the graphical representation, generate a mental
representation by running the motion in your mind. (a) From โ50 s to 0, the
Acela is cruising at a constant positive velocity in the +x direction. Between 0
and 50 s, the constant velocity is maintained for a while and then the train
begins to speed up. Between 50s and 100s, the train speeds up; notice that the
graph is linear in this segment. Between 100 s and 200 s, the train reaches its
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maximum speed of about 175 mi/h. Between 200 s and 300 s, the engineer
applies the brakes: the train is slowing down. The train is moving very slowly
between 300 s and 350 s and then stops at 350 s. After 350 s, the train reverses
direction: the velocity is negative. The train is backing up faster and faster
until the data ends at 400 s.
Categorize Much of the motion of the train does not fit into a simple analysis
model. However, between 50 s and 100 s, the train is well modeled as a
particle under constant acceleration, since the vโt graph is linear.
Analyze (b) Draw a best-fit line between 50 s and 100 s on the graph, taking
advantage of the full vertical extent of the graph:
Now, find the slope of the green line, using its topmost and bottommost
points, reading the coordinates from the graph:
a = slope =
ฮv 200 mi/ h โ ( โ100 mi/ h )
=
= 1.9 mi/ h ยท s
125 s โ ( โ30 s )
ฮt
This result can be converted to metric units:
โ 1 609 m โ โ 1 h โ
a = 1.9 mi/ h ยท s โ
= 0.87 m/ s2
โ
โ
โ
โ 1 mi โ โ 3 600 s โ
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(c) To find the displacement, we need to use the notion in Section 2.9 that the
displacement of a particle is the area under the vโt curve. Because we do not
have a functional description of the velocity of the train, we will need to
approximate the displacement. Letโs approximate the area under the curve
between 0 and 200 s with the following triangle (yellow) and two rectangles
(green):
Find the area of each shape:
Lower rectangle:
ฮx1 = ( 40 mi/ h ) ( 200 s) = 8 000 mi ยท s/ h
Upper rectangle:
ฮx2 = (170 mi/ h โ 40 mi/ h ) (100 s) = 13 000 mi ยท s/ h
Triangle:
ฮx3 = 12 (110 mi/ h ) ( 55 s) = 3 025 mi ยท s/ h
Therefore, the total displacement is
ฮxtotal = ฮx1 + ฮx2 + ฮx3
= 8 000 mi ยท s/ h + 13 000 mi ยท s/ h + 3 025 mi ยท s/ h
= 2.4 ร 10 4 mi ยท s/ h
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This is not a very useful unit! So letโs reconcile the hours and seconds:
โ 1h โ
ฮxtotal = 2.4 ร 10 4 mi ยท s/ h โ
= 6.7 mi
โ 3 600 s โโ
Finalize There are a variety of ways that the area under the curve in part (c)
could be approximated, so the results may vary a bit depending on how this
approximation is made.
Answers: (a) From โ50 s to 0, the Acela is cruising at a constant positive
velocity in the +x direction. Between 0 and 50 s, the constant velocity is
maintained for a while and then the train begins to speed up. Between 50s
and 100s, the train speeds up; notice that the graph is linear in this segment.
Between 100 s and 200 s, the train reaches its maximum speed of about 175
mi/h. Between 200 s and 300 s, the engineer applies the brakes: the train is
slowing down. The train is moving very slowly between 300 s and 350 s and
then stops at 350 s. After 350 s, the train reverses direction: the velocity is
negative. The train is backing up faster and faster until the data ends at 400 s.
(b) 0.87 m/s2 (c) 6.7 mi (answers may vary, depending on estimation from the
graph.)
SOLUTIONS TO END-OF-CHAPTER PROBLEMS
Section 2.1
P2.1
Position, Velocity, and Speed
We assume that you are approximately 2 m tall and that the nerve impulse
travels at uniform speed. The elapsed time is then
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P2.2
We substitute for t in x = 10t2, then use the definition of average velocity:
t (s)
2.00
2.10
3.00
x (m)
40.0
44.1
90.0
(a)
(b)
P2.3
We read the data from the table provided, assume three significant figures of
precision for all the numbers, and use Equation 2.2 for the definition of
average velocity.
(a)
(b)
(c)
Section 2.2
P2.4
Instantaneous Velocity and Speed
We use the definition of average velocity.
(a)
(b)
(c)
To find the average velocity for the round trip, we add the displacement
and time for each of the two halves of the swim:
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(d) The average speed of the round trip is the total distance the athlete
travels divided by the total time for the trip:
P2.5
For average velocity, we find the slope of a secant line running across the graph
between the 1.5-s and 4-s points. Then for instantaneous velocities we think of
slopes of tangent lines, which means the slope of the graph itself at a point.
We place two points on the curve: Point A, at t = 1.5 s, and Point B, at t = 4.0 s,
and read the corresponding values of x.
(a)
At ti = 1.5 s, xi = 8.0 m (Point A)
At tf = 4.0 s, xf = 2.0 m (Point B)
ANS. FIG. P2.5
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(b)
The slope of the tangent line can be found from points C and D. (tC = 1.0
s, xC = 9.5 m) and (tD = 3.5 s, xD = 0),
The negative sign shows that the direction of vxis along the negative x
direction.
(c)
The velocity will be zero when the slope of the tangent line is zero. This
occurs for the point on the graph where x has its minimum value. This is
at
Section 2.3
P2.6
.
Analysis Model: Particle Under Constant Velocity
The trip has two parts: first the car travels at constant speed v1 for distance d,
then it travels at constant speed v2 for distance d. The first part takes the time
interval
(a)
= d/v1, and the second part takes the time interval โt2 = d/v2.
By definition, the average velocity for the entire trip is
where
and
Putting
these together, we have
We know vavg = 30 mi/h and v1 = 60 mi/h.
Solving for v2 gives
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(b)
The average velocity for this trip is
where
so, vavg=
(c)
The average speed for this trip is
d + d = 2d and
where d = d1 + d2 =
so, the average speed is the
same as in part (a): vavg =
P2.7
(a)
The total time for the trip is ttotal = t1 + 22.0 min = t1 + 0.367 h, where t1 is
the time spent traveling at v1 = 89.5 km/h. Thus, the distance traveled is
which gives
or
from which, t1= 2.44 h, for a total time of
(b)
Section 2.5
P2.8
The distance traveled during the trip is
giving
Acceleration
The acceleration is zero whenever the marble is on a horizontal section. The
acceleration has a constant positive value when the marble is rolling on the
20-to-40-cm section and has a constant negative value when it is rolling on the
second sloping section. The position graph is a straight sloping line whenever
the speed is constant and a section of a parabola when the speed changes.
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ANS. FIG. P2.8
P2.9
(a)
In the interval ti = 0 s and tf= 6.00 s, the motorcyclistโs velocity changes
from vi = 0 to vf = 8.00 m/s. Then,
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(b)
Maximum positive acceleration occurs when the slope of the velocitytime curve is greatest, at t = 3 s, and is equal to the slope of the graph,
approximately (6 m/s โ 2 m/s)/(4 s โ 2 s) =
(c)
The acceleration a = 0 when the slope of the velocity-time graph is zero,
which occurs at
, and also for
(d) Maximum negative acceleration occurs when the velocity-time graph
has its maximum negative slope, at t = 8 s, and is equal to the slope of
the graph, approximately
P2.10(a) The graph is shown in ANS. FIG. P2.10 below.
ANS. FIG. P2.10
(b)
At t = 5.0 s, the slope is
.
At t = 4.0 s, the slope is
.
At t = 3.0 s, the slope is
.
At t = 2.0 s, the slope is
.
(c)
(d) The initial velocity of the car was
.
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P2.11
(a)
The area under a graph of a vs. t is equal to the change in velocity, โv.
We can use Figure P2.11 to find the change in velocity during specific
time intervals.
The area under the curve for the time interval 0 to 10 s has the shape of a
rectangle. Its area is
= (2 m/s2)(10 s) = 20 m/s
The particle starts from rest, v0 = 0, so its velocity at the end of the 10-s
time interval is
v = v0 +
= 0 + 20 m/s =
Between t = 10 s and t = 15 s, the area is zero:
= 0 m/s.
Between t = 15 s and t = 20 s, the area is a rectangle:
=
(โ3 m/s2)(5 s) = โ15 m/s.
So, between t = 0 s and t = 20 s, the total area is
= (20 m/s) +
(0 m/s) + (โ15 m/s) = 5 m/s, and the velocity at t = 20 s is
(b)
We can use the information we derived in part (a) to construct a graph
of x vs. t; the area under such a graph is equal to the displacement,
of the particle.
From (a), we have these points (t, v) = (0 s, 0 m/s), (10 s, 20 m/s), (15 s, 20
m/s), and (20 s, 5 m/s). The graph appears below.
The displacements are:
0 to 10 s (area of triangle):
= (1/2)(20 m/s)(10 s) = 100 m
10 to 15 s (area of rectangle):
= (20 m/s)(5 s) = 100 m
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15 to 20 s (area of triangle and rectangle):
= (1/2)[(20 โ 5) m/s](5 s) + (5 m/s)(5 s)
= 37.5 m + 25 m = 62.5 m
Total displacement over the first 20.0 s:
= 100 m + 100 m + 62.5 m = 262.5 m =
Section 2.6
P2.12
Motion Diagrams
(a)
(b)
(c)
(d)
(e)
(f)
One way of phrasing the answer: The spacing of the successive positions
would change with
Another way: The object would move with some combination of the
kinds of motion shown in (a) through (e). Within one drawing, the
acceleration vectors would vary in magnitude and direction.
P2.13
(a)
The motion is fast at first but slowing until the speed is constant. We
assume the acceleration is constant as the object slows.
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(b)
The motion is constant in speed.
(c)
The motion is speeding up, and we suppose the acceleration is constant.
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Section 2.7
P2.14
Analysis Model: Particle Under Constant Acceleration
and x f โ xi = 1.50 ร 10
We have
โ2
m.
(a)
:
(b)
P2.15
In parts (a) โ (c), we use Equation 2.13 to determine the velocity at the times
indicated.
(a)
The time given is 1.00 s after 10:05:00 a.m., so
vf = vi + at = 13.0 m/s + (โ4.00 m/s2)(1.00 s) =
(b)
The time given is 4.00 s after 10:05:00 a.m., so
vf = vi + at = 13.0 m/s + (โ4.00 m/s2)(4.00 s) =
(c)
The time given is 1.00 s before 10:05:00 a.m., so
vf = vi + at = 13.0 m/s + (โ4.00 m/s2)(โ1.00 s) =
(d)
(e)
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P2.16
We think of the plane moving with maximum-size backward acceleration
throughout the landing, so the acceleration is constant, the stopping time a
minimum, and the stopping distance as short as it can be. The negative
acceleration of the plane as it lands can be called deceleration, but it is simpler
to use the single general term acceleration for all rates of velocity change.
(a)
The plane can be modeled as a particle under constant acceleration, with
Given
we use the equation
and solve for t:
(b)
Find the required stopping distance and compare this to the length of
the runway. Taking xi to be zero, we get
or
(c)
P2.17
The stopping distance is greater than the length of the runway;
The velocity is always changing; there is always nonzero acceleration and the
problem says it is constant. So we can use one of the set of equations
describing constant-acceleration motion. Take the initial point to be the
moment when xi= 3.00 cmand vxi = 12.0 cm/s. Also, at t = 2.00 s, xf = โ5.00 cm.
Once you have classified the object as a particle moving with constant
acceleration and have the standard set of four equations in front of you, how
do you choose which equation to use? Make a list of all of the six symbols in
the equations: xi , xf, vxi , vxf, ax, and t. On the list fill in values as above,
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showing that xi, xf, vxi, and t are known. Identify ax as the unknown. Choose
an equation involving only one unknown and the knowns. That is, choose an
equation not involving vxf. Thus we choose the kinematic equation
and solve for ax:
We substitute:
ax =
2[ โ5.00 cm โ 3.00 cm โ(12.0 cm/s)(2.00 s)]
(2.00 s)2
= โ16.0 cm/s 2
P2.18
As in the algebraic solution to Example 2.8, we let t represent the time the
trooper has been moving. We graph
xcar = 45+ 45t
and
xtrooper = 1.5t2
They intersect at
ANS. FIG. P2.18
P2.19
Let the glider enter the photogate with velocity vi and move with constant
acceleration a. For its motion from entry to exit,
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(a)
The speed halfway through the photogate in space is given by
and this is
(b)
The speed halfway through the photogate in time is given by
and this is
P2.20
as determined above.
We ask whether the constant acceleration of the rhinoceros from rest over a
period of 10.0 s can result in a final velocity of 8.00 m/s and a displacement of
50.0 m? To check, we solve for the acceleration in two ways.
P2.21
1)
ti = 0, vi = 0; t = 10.0 s, vf = 8.00 m/s:
2)
ti = 0, xi = 0, vi = 0; t = 10.0 s, xf = 50.0 m:
(a)
Let a stopwatch start from t = 0 as the front end of the glider passes point
A. The average speed of the glider over the interval between t = 0 and
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t = 0.628 s is 12.4 cm/(0.628 s) =
, and this is the instantaneous
speed halfway through the time interval, at t = 0.314 s.
(b)
The average speed of the glider over the time interval between 0.628 +
1.39 = 2.02 s and 0.628 + 1.39 + 0.431 = 2.45 s is
12.4 cm/(0.431 s) = 28.8 cm/s and this is the instantaneous speed at the
instant t = (2.02 + 2.45)/2 = 2.23 s.
Now we know the velocities at two instants, so the acceleration is found
from
(c)
P2.22
Take any two of the standard four equations, such as
Solve one for vxi, and substitute into the other:
vxi = vxf โ axt
Thus
We note that the equation is dimensionally correct. The units are units of
length in each term. Like the standard equation
this
equation represents that displacement is a quadratic function of time.
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P2.23
(a)
For the first car, the speed as a function of time is
For the second car, the speed is
Setting the two expressions equal gives
Solving for t gives
(b)
The first car then has speed
and this is also the constant speed of the second car.
(c)
For the first car, the position as a function of time is
For the second car, the position is
At the point where the cars pass one another, their positions are equal:
rearranging gives
We solve this with the quadratic formula. Suppressing units,
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(d) At t = 0.604 s, the second and also the first carโs position is
At t = 6.90 s, both are at position
(e)
*P2.24
Conceptualize A diagram of the information provided will be helpful. The
diagram below shows three positions of the car and the times at which the car
is at those positions.
Categorize The phrase โslowing down the car uniformlyโ tells us to use the
particle under constant acceleration analysis model for this problem.
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Analyze We have no information about the velocity of the car, so that
suggests we focus on the equation in the particle under constant acceleration
model involving position: x f = xi + vit +
1 2
at . (a) Write the position equation
2
twice, once for the car at pole #2 and once at pole #3, each referencing pole #1
as the initial situation:
x2 = x1 + v1t2 + 12 at22 = 0 + v1t2 + 12 at22 = v1t2 + 12 at22
(1)
x3 = x1 + v1t3 + 12 at32 = 0 + v1t3 + 12 at32 = v1t3 + 12 at32
(2)
Eliminate v1 between Equations (1) and (2) and solve for the acceleration a:
a=
2 ( x3t2 โ x2t3 )
(3)
t32t2 โ t22t3
Substitute numerical values:
2 โก( 80.0 m ) (10.0 s) โ ( 40.0 m ) ( 25.0 s ) โคโฆ
a= โฃ
= โ0.107 m/ s2
2
2
( 25.0 s) (10.0 s) โ (10.0 s) ( 25.0 s)
(b) Solve Equation (1) for the initial speed of the car:
x2 = v1t2 + 12 at22
x2 โ 12 at22
v1 =
t2
โ
(4)
Substitute numerical values:
v1 =
(
)
40.0 m โ 12 โ0.107 m/ s 2 (10.0 s)
(10.0 s)
2
= 4.53 m/ s
(c) Using Equation 2.17 in the particle under constant acceleration model
relating velocity and position, solve for the position at which the velocity goes
to zero:
v = v + 2ax f
2
f
2
1
โ
xf =
v 2f โ v12
2a
=
0 โ ( 4.53 m/ s)
(
2
2 โ0.107 m/ s 2
)
= 96.3 m
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Because the fourth pole would be at x = 120 m, the last pole passed is pole #3.
Finalize Part (c) can also be solved less directly by finding the time at which
the velocity goes to zero and then substituting into the position equation to
find the position at that time. In doing so, you find the time at which the car
stops to be 42.5 s. It might be surprising that the car travels for an additional
42.5 โ 25.0 = 17.5 s beyond pole #3 (longer than between poles #1 and #2, and
between poles #2 and #3), but only moves 96.3 โ 80.0 m = 16.3 m beyond that
pole.
Answers: (a)โ0.107 m/s2(b) 4.53 m/s (c) pole #3 (stops at x = 96.3 m at t = 42.4 s
Section 2.8
P2.25
Freely Falling Objects
The bill starts from rest, vi = 0, and falls with a downward acceleration of 9.80
m/s2 (due to gravity). For an average human reaction time of about 0.20 s, we
can find the distance the bill will fall:
The bill falls about 20 cmโthis distance is about twice the distance between
the center of the bill and its top edge, about 8 cm. Thus
P2.26
We can solve (a) and (b) at the same time by assuming the rock passes the top
of the wall and finding its speed there. If the speed comes out imaginary, the
rock will not reach this elevation.
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which gives
.
(c)
The rock travels from yi = 3.65 m to yf = 1.55 m. We find the final speed of
the rock thrown down:
which gives
The change in speed of the rock thrown down is
(d) The magnitude of the speed change of the rock thrown up is
This
with 2.39 m/s.
(e)
P2.27
We are given the height of the helicopter: y = h = 3.00t3.
At t = 2.00 s, y = 3.00(2.00 s)3 = 24.0 m and
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If the helicopter releases a small mailbag at this time, the mailbag starts its
free fall with velocity 36.0 m/s upward. The equation of motion of the
mailbag is
Setting yf = 0, dropping units, and rearranging the equation, we have
4.90t2 โ 36.0t โ 24.0 = 0
We solve for t using the quadratic formula:
Since only positive values of t count, we find
P2.28
The falling ball moves a distance of (15 m โ h) before they meet, where h is the
height above the ground where they meet. We apply
to the falling ball to obtain
or
Applying
[1]
to the rising ball gives
[2]
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Combining equations [1] and [2] gives
or
P2.29
We model the keys as a particle under the constant free-fall acceleration. Take
the first studentโs position to be
and the second studentโs position to be
We are given that the time of flight of the keys is t = 1.50 s, and
.
(a)
We choose the equation
to connect the data and the
unknown.
We solve:
and substitute:
(b)
The velocity at any time t > 0 is given by vyf = vyi + ayt.
Therefore, at t =1.50 s,
The negative sign means that the keys are moving downward just before
they are caught.
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P2.30
(a)
The keys, moving freely under the influence of gravity (a = โg), undergo
a vertical displacement of โy = +h in time t. We use
to find
the initial velocity as
(b)
We find the velocity of the keys just before they were caught (at time t)
using v = vi + at:
*P2.31
Conceptualize This is a simple situation in which an object is thrown straight
upward and is in free fall after being thrown. We wish to relate the speed of
the throw and the height of the projectile.
Categorize We recognize a simple one-dimensional free-fall problem, in
which we use the particle under constant acceleration model for the motion of
the box.
Analyze (a) From the particle under constant acceleration model, chose the
equation that relates position and equation (Equation 2.17), noting that the
motion is in the y direction:
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(1)
Consider the point at which the box reaches its highest position. Note that the
velocity of the box at that position is vyf = 0, and define the point at which the
box is thrown as the origin, so that yi = 0. Also assign the acceleration to be
that due to gravity, so that Equation (1) becomes
(2)
Find the highest position for the box if thrown upward at the speed of 20.0
m/s found in the demonstration:
Because this position is higher than the bottom of the window in which the
alleged accomplice caught the box, the action of throwing the box to the
accomplice is possible.
(b) The defense attorney could argue as follows: The demonstration involved
throwing a baseball horizontally. It is more difficult to throw something
upward, so the throwing speed could be in error. You might counter-argue
that the throwing speed could actually be higher because the defendant was
attempting to minimize his throwing speed during the demonstration in
order to show that the feat could not be done.
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Answers: (a) The box could reach the window according to the data
provided.(b) Answers will vary.
Section 2.9
P2.32
(a)
Kinematic Equations Derived from Calculus
See the x vs. t graph on the top panel of ANS. FIG. P2.32.Choosex = 0 at t
= 0.
(b)
See the a vs. t graph at the bottom right.
For 3 <t< 5 s, a = 0.
ANS. FIG. P2.32
At the points of inflection, t = 3 and 5 s, the slope of the velocity curve
changes abruptly, so the acceleration is not defined.
(c)
For 5 s <t< 9 s,
(d) The average velocity between t = 5 and 7 s is
vavg = (8 m/s +0)/2 = 4 m/s
At
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(e)
The average velocity between t = 5 and 9 s is
vavg = [(8 m/s) + (โ8 m/s)]/2 = 0 m/s
At t = 9 s, x = 28 m + (0 m/s)(1 s) =
P2.33
This is a derivation problem. We start from basic definitions. We are given J =
dax/dt= constant, so we know that dax = Jdt.
(a)
Integrating from the โinitialโ moment when we know the acceleration to
any later moment,
Therefore,
From ax = dvx/dt,
dvx = axdt.
Integration between the same two points tells us the velocity as a
function of time:
From vx = dx/dt, dx = vxdt. Integrating a third time gives us x(t):
and
(b)
Squaring the acceleration,
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Rearranging,
The expression for vxwas
So
and by substitution
Additional Problems
P2.34
(a)
Area A1 is a rectangle. Thus, A1 = hw = vxit.
Area A2 is triangular. Therefore,
The total area under the curve is
and since vx โ vxi = axt,
(b)
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P2.35
(a)
From
the insectโs velocity after straightening its legs is
(b)
The time to reach this velocity is
(c)
The upward displacement of the insect between when its feet leave the
ground and its speed is momentarily zero is
P2.36
Take downward as the positive y direction.
(a)
While the woman was in free fall,
and we take
Thus,
giving
(b)
Her velocity just before impact is:
While crushing the box,
,
, and
Therefore,
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or
(c)
Time to crush box:
or
P2.37
(a)
In order for the trailing athlete to be able to catch the leader, his speed
(v1) must be greater than that of the leading athlete (v2), and the distance
between the leading athlete and the finish line must be great enough to
give the trailing athlete sufficient time to make up the deficient distance,
d.
(b)
During a time interval t the leading athlete will travel a distance d2 = v2t
and the trailing athlete will travel a distance d1 = v1t. Only when d1 = d2 +
d (where d is the initial distance the trailing athlete was behind the
leader) will the trailing athlete have caught the leader. Requiring that
this condition be satisfied gives the elapsed time required for the second
athlete to overtake the first:
d1 = d2 + d
or
v1t = v2t + d
giving
(c)
In order for the trailing athlete to be able to at least tie for first place, the
initial distance D between the leader and the finish line must be greater
than or equal to the distance the leader can travel in the time t calculated
above (i.e., the time required to overtake the leader). That is, we must
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require that
P2.38
For the collision not to occur, the front of the passenger train must not have a
position that is equal to or greater than the position of the back of the freight
train at any time. We can write expressions of position to see whether the
front of the passenger car (P) meets the back of the freight car (F) at some
time.
Assume at t = 0, the coordinate of the front of the passenger car is
xPi = 0; and the coordinate of the back of the freight car is xFi = 58.5 m. At later
time t, the coordinate of the front of the passenger car is
and the coordinate of the back of the freight car is
Setting these expression equal to each other gives
or
after simplifying and suppressing units.
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We do not have to solve this equation, we just want to check if a solution
exists; if a solution does exist, then the trains collide. A solution does exist:
P2.39
We have constant-acceleration equations to apply to the two cars separately.
(a)
Let the times of travel for Hannah and Stan be tKand tS, where
tS = tK + 1.00 s
Both start from rest (vxi,K= vxi,S = 0), so the expressions for the distances
traveled are
and
When Hannah overtakes Stan, the two distances will be equal. Setting xK
= xSgives
This we simplify and write in the standard form of a quadratic as
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We solve using the quadratic formula
, suppressing
units, to find
Only the positive root makes sense physically, because the overtake
point must be after the starting point in time.
P2.40
(b)
Use the equation from part (a) for distance of travel,
(c)
Remembering that vxi,K= vxi,S= 0, the final velocities will be:
We translate from a pictorial representation through a geometric model to a
mathematical representation by observing that the distances x and y are
always related by x2+y2 = L2 .
(a)
Differentiating this equation with respect to time, we have
Now the unknown velocity of B is
and
so the differentiated equation becomes
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But
so
(b)
P2.41
The average speed of every point on the train as the first car passes Lisa is
given by:
The train has this as its instantaneous speed halfway through the 1.50-s time.
Similarly, halfway through the next 1.10 s, the speed of the train is
. The time required for the speed to change from 5.73 m/s
to 7.82 m/s is
so the acceleration is:
Challenge Problems
P2.42
(a)
The factors to consider are as follows. The red bead falls through a
greater distance with a downward acceleration of g. The blue bead
travels a shorter distance, but with acceleration of
A first guess
would be that the blue bead โwins,โ but not by much. We do note,
however, that points
with
(b)
,
, and
are the vertices of a right triangle
as the hypotenuse.
The red bead is a particle under constant acceleration. Taking
downward as the positive direction, we can write
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as
which gives
(c)
The blue bead is a particle under constant acceleration, with
Taking the direction along L as the positive direction, we can write
as
which gives
(d) For the two beads to reach point
simultaneously, tR = tB. Then,
Squaring both sides and cross-multiplying gives
or
We note that the angle between chords
(e)
P2.43
and
that the angle between chords
and
and the beads arrive at point
simultaneously.
is
is
so
Then,
Once we recognize that the two rods form one side and the hypotenuse
of a right triangle with ฮธ as its smallest angle, then the result becomes
obvious.
Consider the runners in general. Each completes the race in a total time
so each
interval T. Each runs at constant acceleration a for a time interval
covers a distance (displacement)
final speed (velocity)
where they eventually reach a
after which they run at this constant speed for
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the remaining time
until the end of the race, covering distance
The total distance (displacement) each covers is
the same:
so
where
(a)
and T = 10.4 s.
For Laura (runner 1),
= 2.00 s:
a1 = (100 m)/(18.8 s2) =
For Healan (runner 2),
= 3.00 s:
a2 = (100 m)/(26.7 s2) =
(b)
Laura (runner 1): v1 = a1
Healan (runner 2): v2 = a2
(c)
=
=
The 6.00-s mark occurs after either time interval
above, each has covered the distance
From the reasoning
where t = 6.00 s.
Laura (runner 1):
= 53.19 m
Healan (runner 2):
= 50.56 m
(d) Laura accelerates at the greater rate, so she will be ahead of Healen at,
and immediately after, the 2.00-s mark. After the 3.00-s mark, Healan is
travelling faster than Laura, so the distance between them will shrink. In
the time interval
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During that time interval, the distance between them (the position of
Laura relative to Healan) is
because Laura has ceased to accelerate but Healan is still accelerating.
Differentiating with respect to time, (and doing some simplification), we
can solve for the time t when D is an maximum:
which gives
Substituting this time back into the expression for D, we find that D =
4.47 m, that is, Laura ahead of Healan by
*P2.44
Conceptualize You may have had this experience when driving. A bicycle
can exhibit a high acceleration, so it can pull ahead of the car when leaving a
traffic light. But the maximum speed of the bicycle is less than that of the car,
so the car will eventually catch up and pass the bicycle. A graphical
representation may be of some help here:
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Because of the higher acceleration of the bicycle, the line representing its velocity is
steeper in slope than that of the car. But it soon reaches its maximum speed of 20.0
mi/h, and the line for its velocity changes to horizontal. The slope of the line
representing the acceleration of your car is less steep, but the acceleration lasts for a
longer time, so that eventually your car is traveling faster than the bicycle.
Categorize The phrase โconstant accelerationโ tells us to use the particle under
constant acceleration analysis model for this problem. Once the vehicles reach the
indicated speeds, they travel at constant velocity, so we will also need to employ the
particle under constant velocity analysis model.
Analyze Assuming that your car catches up with the bicycle before reaching its
maximum speed, the motion of your car is described completely during the time
interval of interest by the particle under constant acceleration model. Therefore, its
position is given by
xcar = xcar ,i + vcar ,it + 12 acart 2 = 12 acart 2 (1)
where we have incorporated the fact that the car begins from rest (vcar,i = 0), and have
defined the position of the traffic light as the origin (xcar,i = 0). The position of the
bicycle is described by the particle under constant acceleration model until it reaches
its maximum speed and by the particle under constant velocity model thereafter.
Therefore, knowing that the bicycle also begins from the origin at rest,
where t1 is the time at which the bicycle reaches its maximum speed and vbicyle,max is
that maximum speed. As noted, the bicycle maintains that speed thereafter. The time
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at which the bicycle has a given speed is provided by the velocity equation in the
particle under constant acceleration model:
vbicycle = vbicycle ,i + abicyclet = 0 + abicyclet
โ t=
(4)
vbicycle
abicycle
Evaluate this time when the bicycle first reaches its maximum speed:
t1 =
(5)
vbicycle,max
abicycle
(a) To find the time interval during which the bicycle is ahead of the car, we find the
time at which the car and the bicycle are at the same position by using Equations (1)
and (3):
xbicycle = xcar
โ
(
1
a
t 2 + vbicycle,max t โ t1
2 bicycle 1
) = 12 a t
2
car
Rearranging this equation gives the standard form of a quadratic equation:
(
1
a t 2 โ vbicycle,max t โ 12 abicyclet12 โ vbicycle,maxt1
2 car
) = 0 (6)
Applying the quadratic formula, we can solve for t as follows:
t=
=
vbicycle,max ยฑ
(
โvbicycle,max
) โ 4 ( a ) โกโฃโ ( a
2
1
2
1
2
car
)
t โ vbicycle,maxt1 โค
โฆ
2
bicycle 1
acar
vbicycle,max ยฑ
( โv
) + 2a ( a
2
bicycle,max
car
1
2
t โ vbicycle,maxt1
2
bicycle 1
)
acar
Finally, substitute for t1:
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(7)
vbicycle,max ยฑ
( โv
bicycle,max
t=
=
)
2
2
โก
โ
โ
โ vbicycle,max โ โค
v
bicycle,max
1
+ 2acar โข 2 abicycle โ
โ โ vbicycle,max โ a
โโฅ
โข
โ abicycle โ
โ bicycle โ โฅ
โฃ
โฆ
acar
โ
acar โ
2
vbicycle,max ยฑ vbicycle,max
1
โ
โ
abicycle โโ
โ
acar
Substitute numerical values:
20.0 mi/ h ยฑ
t=
โก
โโค
โ โฆ
โ
โ
9.00 mi/ h/ s
( 20.0 mi/ h ) โข1โ โ 13.0
โฅ
mi/ h/ s โ
2
โฃ
9.00 mi/ h/ s
= 3.45 s
(The other root is t = 0.990 s, but this turns out to be less than t1, so it is not a realistic
solution.)
(b) The speed of the your car at any time is found from the particle under constant
acceleration model:
vcar = vcar ,i + acart
(8)
The maximum separation occurs when the speeds of both vehicles are the same,
because after that instant the car will be moving faster than the bicycle, so the
separation distance will decrease. Therefore, set the speed of the car equal to the
speed of the bicycle in Equation (8) and find the time t2 at which the two objects have
the same speed:
vcar = vbicycle,max = vcar ,i + acart2
โ t2 =
vbicycle,max
(9)
acar
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At any time after the bicycle reaches its maximum speed, the separation distance
between the vehicles is found by subtracting Equation (1) from Equation (3):
xbicycle โ xcar = 12 abicyclet12 + vbicycle,max ( t โ t1 ) โ 12 acart 2
(10)
Evaluate the separation distance at the time found in Equation (9):
xbicycle โ xcar = 12 abicyclet12 + vbicycle,max ( t2 โ t1 ) โ 12 acart12
2
โ vbicycle,max โ
โ vbicycle,max vbicycle,max โ
= 12 abicycle โ
+
v
โ
โ
bicycle,max โ
abicycle โโ
โ abicycle โ
โ acar
โ vbicycle,max โ
โ 12 acar โ
โโ
โ a
2
car
(
= vbicycle,max
)
2
โก 1
โ 1
1 โ
1 โค
+โ
โ
โข
โฅ
โโ
โขโฃ 2abicycle โ acar abicycle โ 2acar โฅโฆ
Substitute numerical values:
1
โก
โค
โข 2 ( 13.0 mi/ h/ s )
โฅ
โข
โฅ
โข
โ
โโฅ
1
1
2
xbicycle โ xcar = ( 20.0 mi/ h ) โข
+โ
โ
โฅ
โ 9.00 mi/ h/ s 13.0 mi/ h/ s โโ โฅ
โข
โข
โฅ
1
โ
โข
โฅ
2 ( 9.00 mi/ h/ s )
โฃ
โฆ
โ 5 280 ft โ โ 1 h โ
= 10.0 ft
= 6.84 mi ยท s/ h โ
โ mi โโ โโ 3 600 s โโ
Finalize It would be helpful to graph the positions of the two vehicles against time as
shown below:
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This graph shows both the time at which the car passes the bicycle and the time at
which the maximum separation occurs between the vehicles. Notice that the bicycle
reaches its maximum speed at about 1.5 s, but the car is still accelerating at 4 seconds
(the blue line is still curved). The black dot represents the instant that the bicycle
stops accelerating. The corresponding black dot for the car would be to the right of
the graph as drawn, at about 5.6 s.]
Answers: (a) 3.45 s (b) 10.0 ft
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ANSWERS TO QUICK-QUIZZES
1. (b)
2. (c)
3. (b)
4. False.
5. (b)
6. (c)
7. (a)โ(e), (b)โ(d), (c)โ(f)
8. (i) (e) (ii) (d)
ANSWERS TO EVEN-NUMBERED PROBLEMS
P2.2
(a) 50.0 m/s; (b) 41.0 m/s
P2.4
(a) +L/t1; (b) โL/t2; (c) 0; (d) 2L/t1+ t2
P2.6
(a) 20 mi/h; (b) 0; (c) 30 mi/h
P2.8
See graphs in P2.8.
P2.10
(a) See ANS. FIG. P2.10; (b) 23 m/s, 18 m/s, 14 m/s, and 9.0 m/s; (c) 4.6 m/s2;
(d) zero
P2.12
(aโe) See graphs in P2.12; (f) with less regularity
P2.14
(a) 4.98ร10-19s
(b) 1.20ร1015m/s2
P2.16
(a) 20.0s (b) 1000m
(c) the plane cannot land
P2.18
31s
P2.20
The accelerations do not match, therefore the situation is impossible.
ยฉ 2019 Cengage Learning, Inc. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
P2.22
The equation represents that displacement is a quadratic function of time.
P2.24
(a)โ0.107 m/s2(b) 4.53 m/s (c) pole #3 (stops at x = 96.3 m at t = 42.4 s
P2.26
(a and b) The rock does not reach the top of the wall with vf = 3.69 m/s; (c)
2.39m/s; (d) does not agree; (e) The average speed of the upward-moving rock
is smaller than the downward moving rock.
P2.28
0.60 s
P2.30
(a)
P2.32
(a) See graphs in P2.32; (b) See graph in P2.32; (c) โ4 m/s2; (d) 32 m; (e) 28 m
P2.34
(a)
P2.36
(a) 96.0 ft/s; (b)
P2.38
The trains do collide.
P2.40
(a) vB= (1/tan u)v (b) The velocity vB starts off larger than v for small angles
(b)
(b) The displacement is the same result for the total area.
; (c)
theta and then decreases, approaching zero as theta approaches 90ยฐ.
P2.42
(a) The red bead falls through a greater distance with a downward
acceleration of g. The blue bead travels a shorter distance, but with
acceleration of
A first guess would be that the blue bead โwins,โ but
not by much. (b)
(c)
(d) the beads arrive at point ยฉ
simultaneously; (e) Once we recognize that the two rods form one side and
the hypotenuse of a right triangle with ฮธ as its smallest angle, then the result
becomes obvious.
P2.44
(a) 3.45 s; (b) 10.0 ft.
ยฉ 2019 Cengage Learning, Inc. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

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