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Topic 2
38
Topic 2
Motion in One Dimension
QUICK QUIZZES
2.1
(a) 200 yd
(b)
0
(c)
0
2.2
(a) False. The car may be slowing down, so that the direction of its
acceleration is opposite the direction of its velocity.
(b) True. If the velocity is in the direction chosen as negative, a positive
acceleration causes a decrease in speed.
(c) True. For an accelerating particle to stop at all, the velocity and
acceleration must have opposite signs, so that the speed is decreasing.
If this is the case, the particle will eventually come to rest. If the
acceleration remains constant, however, the particle must begin to
move again, opposite to the direction of its original velocity. If the
particle comes to rest and then stays at rest, the acceleration has
become zero at the moment the motion stops. This is the case for a
braking carโthe acceleration is negative and goes to zero as the car
comes to rest.
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Topic 2
2.3
39
The velocity-vs-time graph (a) has a constant slope, indicating a constant
acceleration, which is represented by the acceleration-vs.-time graph (e).
Graph (b) represents an object whose speed always increases, and does so
at an ever-increasing rate. Thus, the acceleration must be increasing, and
the acceleration-vs-time graph that best indicates this behaviour is (d).
Graph (c) depicts an object which first has a velocity that increases at a
constant rate, which means that the objectโs acceleration is constant. The
motion then changes to one at constant speed, indicating that the
acceleration of the object becomes zero. Thus, the best match to this
situation is graph (f).
2.4
Choice (b). According to graph b, there are some instants in time when the
object is simultaneously at two different x-coordinates. This is physically
impossible.
2.5
(a) The blue graph of Figure 2.14b best shows the puckโs position as a
function of time. As seen in Figure 2.14a, the distance the puck has
traveled grows at an increasing rate for approximately three time
intervals, grows at a steady rate for about four time intervals, and
then grows at a diminishing rate for the last two intervals.
(b) The red graph of Figure 2.14c best illustrates the speed (distance
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Topic 2
40
traveled per time interval) of the puck as a function of time. It shows
the puck gaining speed for approximately three time intervals,
moving at constant speed for about four time intervals, then slowing
to rest during the last two intervals.
(c) The green graph of Figure 2.14d best shows the puckโs acceleration as
a function of time. The puck gains velocity (positive acceleration) for
approximately three time intervals, moves at constant velocity (zero
acceleration) for about four time intervals, and then loses velocity
(negative acceleration) for roughly the last two time intervals.
2.6
Choice (e). The acceleration of the ball remains constant while it is in the
air. The magnitude of its acceleration is the free-fall acceleration, g = 9.80
m/s2.
2.7
Choice (c). As it travels upward, its speed decreases by 9.80 m/s during
each second of its motion. When it reaches the peak of its motion, its
speed becomes zero. As the ball moves downward, its speed increases by
9.80 m/s each second.
2.8
Choices (a) and (f). The first jumper will always be moving with a higher
velocity than the second. Thus, in a given time interval, the first jumper
covers more distance than the second, and the separation distance
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Topic 2
41
between them increases. At any given instant of time, the velocities of the
jumpers are definitely different, because one had a head start. In a time
interval after this instant, however, each jumper increases his or her
velocity by the same amount, because they have the same acceleration.
Thus, the difference in velocities stays the same.
ANSWERS TO EVEN NUMBERED CONCEPTUAL QUESTIONS
2.2
Yes. The particle may stop at some instant, but still have an acceleration,
as when a ball thrown straight up reaches its maximum height.
2.4
(a) No. They can be used only when the acceleration is constant.
(b) Yes. Zero is a constant.
2.6
(a) In Figure (c), the images are farther apart for each successive time
interval. The object is moving toward the right and speeding up. This
means that the acceleration is positive in Figure (c).
(b) In Figure (a), the first four images show an increasing distance
traveled each time interval and therefore a positive acceleration.
However, after the fourth image, the spacing is decreasing, showing
that the object is now slowing down (or has negative acceleration).
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Topic 2
42
(c) In Figure (b), the images are equally spaced, showing that the object
moved the same distance in each time interval. Hence, the velocity is
constant in Figure (b).
2.8
(a) At the maximum height, the ball is momentarily at rest (i.e., has zero
velocity). The acceleration remains constant, with magnitude equal to
the free-fall acceleration g and directed downward. Thus, even
though the velocity is momentarily zero, it continues to change, and
the ball will begin to gain speed in the downward direction.
(b) The acceleration of the ball remains constant in magnitude and
direction throughout the ballโs free flight, from the instant it leaves
the hand until the instant just before it strikes the ground. The
acceleration is directed downward and has a magnitude equal to the
freefall acceleration g.
2.10
Once the ball has left the thrower’s hand, it is a freely falling body with a
constant, nonzero, acceleration of a = โg. Since the acceleration of the ball
is not zero at any point on its trajectory, choices (a) through (d) are all
false and the correct response is (e).
2.12
The initial velocity of the car is v0 = 0 and the velocity at the time t is v.
The constant acceleration is therefore given by
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Topic 2
43
a=
ฮv v โ v0 v โ 0 v
=
=
=
ฮt
t
t
t
and the average velocity of the car is
v=
( v + v0 ) = ( v + 0 ) = v
2
2
2
The distance traveled in time t is ฮx = vt = vt/2. In the special case where
a = 0 (and hence v = v0 = 0), we see that statements (a), (b), (c), and (d) are
all correct. However, in the general case (a โ 0, and hence v โ 0) only
statements (b) and (c) are true. Statement (e) is not true in either case.
ANSWERS TO EVEN NUMBERED PROBLEMS
2.2
(a)
2 ร 104 mi
(b)
ฮx/2RE = 2.4
2.4
(a)
8.33 yards/s
(b)
2.78 yards/s
2.6
(a)
5.00 m/s
(b)
1.25 m/s
(d)
โ3.33 m/s
(e)
0
(a)
+4.0 m/s
(b)
โ0.50 m/s
(d)
0
(a)
2.3 min
(b)
64 mi
2.8
2.10
(c)
โ2.50 m/s
(c)
โ1.0 m/s
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Topic 2
2.12
44
(a)
L/t1
(b)
โL/t2
(d)
2L/(t1 + t2)
2.14
(a)
2.16
2.18
(c)
0
1.3 ร 102 s
(b)
13 m
(a)
37.1 m/s
(b)
1.30 ร 10โ5 m
(a)
Some data points that can be used to plot the graph are as given
below:
x (m)
5.75
16.0
35.3
68.0
119
192
t (s)
1.00
2.00
3.00
4.00
5.00
6.00
(b)
41.0 m/s, 41.0 m/s, 41.0 m/s
(c)
17.0 m/s, much smaller than the instantaneous velocity at t = 4.00 s
2.20
(a)
2.00 m/s, 5.0 m/s (b)
2.22
0.391 s
2.24
(i)
(a)
0
(b)
1.6 m/s2
(c)
0.80 m/s2
(ii)
(a)
0
(b)
1.6 m/s2
(c)
0
2.26
263 m
The curves intersect at t = 16.9 s.
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Topic 2
45
2.28
a = 2.74 ร 105 m/s2 = (2.79 ร 104)g
2.30
(a)
(c)
a = (v 2f โ vi2 )/ ( 2 ฮx)
(b)
13.5 m (c)
(b)
v 2f = vi2 + 2a(ฮx)
(e)
8.00 s
(a)
13.5 m
(d)
22.5 m
(a)
20.0 s
(b)
No, it cannot land safely on the 0.800 km runway.
2.36
(a)
5.51 km
(b)
20.8 m/s, 41.5 m/s, 20.8 m/s, 38.7 m/s
2.38
(a)
107 m
(b)
1.49 m/s2
2.40
(a)
v = a1t1
(b)
ฮx =
(c)
ฮxtotal =
1 2
1
a1t1 + a1t1t 2 + a2t 22
2
2
2.32
2.34
(d)
1.25 m/s2
13.5 m
1 2
a1t1
2
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Topic 2
46
2.42
8.9 months
2.44
29.1 s
2.46
1.79 s
2.48
(a)
Yes.
(c)
๏ฒ
ฮv downward = 2.39 m/s
(d)
!
No, ฮv upward = 3.71 m/s. The two rocks have the same acceleration,
(b)
vtop = 3.69 m/s
but the rock thrown downward has a higher average speed between
the two levels, and is accelerated over a smaller time interval.
2.50
(a)
21.1 m/s
(b)
19.6 m
2.52
(a)
v = | โv0 โ gt| = |v0 + gt|
(c)
v = v0 โ gt , d =
2.54
(a)
29.4 m/s
(b)
44.1 m
2.56
(a)
โ202 m/s2
(b)
198 m
2.58
(a)
4.53 s
(b)
14.1 m/s
2.60
8.4 m/s
2.62
See Solutions Section for Motion Diagrams.
(c)
1.81 m/s, 19.6 m
(b)
d=
1 2
gt
2
1 2
gt
2
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Topic 2
47
2.64
(a)
v = v02 + 2gh
(b)
ฮt = 2v0/g
2.66
(a)
2.45 ร 10โ2 m
(b)
4.67 ร 10โ2 s
2.68
(a)
3.00 s
(b)
v0,2 = โ15.2 m/s
(c)
v1 = โ31.4 m/s, v2 = โ34.8 m/s
(a)
2.2 s
2.70
(b)
โ21 m/s
(c) 2.3 s
PROBLEM SOLUTIONS
2.1
We assume that you are approximately 2 m tall and that the nerve
impulse travels at uniform speed. The elapsed time is then
ฮt =
2.2
ฮx
2m
=
= 2 ร 10 โ2 s = 0.02 s
v 100 m/s
(a) At constant speed, c = 3 ร 108 m/s, the distance light travels in 0.1 s is
ฮx = c(ฮt) = (3 ร 10 8 m/s)(0.1 s)
โ 1 mi โ โ 1 km โ
= (3 ร 10 7 m ) โ
= 2 ร 10 4 mi
โ 1.609 km โโ โโ 10 3 m โโ
(b) Comparing the result of part (a) to the diameter of the Earth, DE, we
find
ฮx
ฮx
3.0 ร 10 7 m
=
=
โ 2.4
DE 2RE 2(6.38 ร 10 6 m )
(with RE = Earth’s radius)
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Topic 2
2.3
48
Distances traveled between pairs of cities are
ฮx1 = v1(ฮt1) = (80.0 km/h)(0.500 h) = 40.0 km
ฮx2 = v2(ฮt2) = (100.0 km/h)(0.200 h) = 20.0 km
ฮx3 = v3(ฮt3) = (40.0 km/h)(0.750 h) = 30.0 km
Thus, the total distance traveled is ฮx = (40.0 + 20.0 + 30.0) km = 90.0 km,
and the elapsed time is ฮt = 0.500 h + 0.200 h + 0.750 h + 0.250 h = 1.70 h.
(a)
v=
ฮx 90.0 km
=
= 52.9 km/h
ฮt
1.70 h
(b) ฮx = 90.0 km (see above)
2.4
(a) The player runs 100 yards from his own goal line to the opposing
teamโs goal line. Then he runs an additional 50 yards back to the fiftyyard line, all in 18.0 s. Substitute values into the definition of average
speed to find
Average speed =
path length 100 yards + 50 yards
=
elapsed time
18.0 s
= 8.33 yards/s
(b) After returning to the fifty-yard line, the playerโs displacement is ฮx =
xf โ xi = 50.0 yards โ 0 yards = 50.0 yards. Substitute values into the
definition of average velocity to find
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Topic 2
49
v=
ฮx 50.0 yards
=
ฮt
18.0 s
= 2.78 yards/s
2.5
(a) Boat A requires 1.0 h to cross the lake and 1.0 h to return, total time
2.0 h. Boat B requires 2.0 h to cross the lake at which time the race is
over. Boat A wins, being 60 km ahead of B when the race ends.
(b) Average velocity is the net displacement of the boat divided by the
total elapsed time. The winning boat is back where it started, its
displacement thus being zero, yielding an average velocity of zero.
2.6
The average velocity over any time interval is
v=
ฮx x f โ xi
=
ฮt t f โ ti
v=
ฮx 10.0 m โ 0
=
= 5.00 m/s
ฮt 2.00 s โ 0
(b) v =
ฮx 5.00 m โ 0
=
= 1.25 m/s
ฮt
4.00 s โ 0
(a)
v=
ฮx 5.00 m โ 10.0 m
=
= โ2.50 m/s
ฮt
4.00 s โ 2.00 s
(d) v =
ฮx โ5.00 m โ 5.00 m
=
= โ3.33 m/s
ฮt
7.00 s โ 4.00 s
(c)
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Topic 2
50
(e)
2.7
v=
ฮx x2 โ x1
0โ0
=
=
= 0
ฮt t 2 โ t1 8.00s โ 0
(a)
โ 1h โ
Displacement = ฮx = (8.50 km/h)(35.0 min) โ
+ 130 km = 180 km
โ 60.0 min โโ
(b) The total elapsed time is
โ 1h โ
ฮt = (35.0 min + 15.0 min) โ
+ 2.00 h = 2.83 h
โ 60.0 min โโ
so,
2.8
v=
ฮx 180 km
=
= 63.6 km
ฮt
2.84 h
The average velocity over any time interval is
(a)
v=
ฮx x f โ xi
=
ฮt t f โ ti
v=
ฮx 4.0 m โ 0
=
= +4.0 m / s
ฮt 1.0 s โ 0
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Topic 2
51
(b) v =
ฮx โ2.0 m โ 0
=
= โ0.50 m / s
ฮt
4.0 s โ 0
v=
ฮx
0 โ 4.0 m
=
= โ1.0 m / s
ฮt 5.0 s โ 1.0 s
(c)
(d) v =
2.9
ฮx
0โ0
=
= 0
ฮt 5.0 s โ 0
The plane starts from rest (v0 = 0) and maintains a constant acceleration of
a = +1.3 m/s2. Thus, we find the distance it will travel before reaching the
required takeoff speed (v = 75 m/s), from v 2 = v02 + 2a(ฮx) , as
ฮx =
v 2 โ v02
(75 m/s)2
=
= 2.2 ร 10 3 m = 2.2 km
2
2a
2(1.3 m/s )
Since this distance is less than the length of the runway,
the plane takes off safely.
2.10
(a) The time for a car to make the trip is t =
ฮx
. Thus, the difference in
v
the times for the two cars to complete the same 10 mile trip is
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Topic 2
52
ฮt = t1 โ t 2 =
ฮx ฮx โ 10 mi
10 mi โ โ 60 min โ
โ
=โ
โ
= 2.3 min
v1 v2 โ 55 mi/h 70 mi/h โโ โโ 1 h โโ
(b) When the faster car has a 15.0 min lead, it is ahead by a distance
equal to that traveled by the slower car in a time of 15.0 min. This
distance is given by ฮx1 = v1(ฮt) = (55 mi/h)(15 min).
The faster car pulls ahead of the slower car at a rate of
vrelative = 70 mi/h โ 55 mi/h = 15 mi/h
Thus, the time required for it to get distance ฮx1 ahead is
ฮt =
ฮx1
(55 mi/h)(15 min)
=
= 15 min
vrelative
15.0 mi/h
Finally, the distance the faster car has traveled during this time is
โ 1h โ
ฮx2 = v2 (ฮt) = (70 mi/h)(55 min) โ
= 64 mi
โ 60 min โโ
2.11
(a) From v f + vi + 2a(ฮx) , with vi = 0, vf = 72 km/h, and ฮx = 45 m, the
2
2
acceleration of the cheetah is found to be
โกโ
km โ โ 10 3 m โ โ 1 h โ โค
72
โข
โฅโ0
h โโ โโ 1 km โโ โโ 3600 s โโ โฆ
v 2f โ vi2 โฃโโ
a=
=
= 4.4 m/s 2
2(ฮx)
2(45 m)
(b) The cheetahโs displacement 3.5 s after starting from rest is
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Topic 2
53
1
1
ฮx = vi t + at 2 = 0 + (4.4 m/s 2 )(3.5 s) = 27 m
2
2
2.12
v1 =
(ฮx)1 +L
=
= +L /t1
(ฮt)1
t1
(b) v2 =
(ฮx)2 โL
=
= โL /t 2
(ฮt)2
t2
(a)
(c)
vtotal =
(ฮx)total (ฮx)1 + (ฮx)2 +L โ L
0
=
=
=
= 0
(ฮt)total
t1 + t 2
t1 + t 2
t1 + t 2
(d)
(ave speed)trip =
2.13
+L + โL
totaldistance traveled (ฮx)1 + (ฮx)2
2L
=
=
=
(ฮt)total
t1 + t 2
t1 + t 2
t1 + t 2
(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 ฮx = v1t1 = vt total , which gives
(89.5 km/h)t1 = (77.8 km/h)(t1 + 0.367 h) = (77.8 km/h)t1 + 28.5 km
or, (89.5 km/h โ 77.8 km/h)t1 = 28.5 km
From which, t1 = 2.44 h for a total time of ttotal = t1 + 0.367 h = 2.81 h
(b) The distance traveled during the trip is ฮx = v1t1 = vt total , giving
ฮx = vt total = (77.8 km/h)(2.81 h) = 219 km
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Topic 2
2.14
54
(a) At the end of the race, the tortoise has been moving for time t and the
hare for a time t โ 2.0 min = t โ 120 s. The speed of the tortoise is vi =
0.100 m/s, and the speed of the hare is vh = 20 vt = 2.0 m/s. The tortoise
travels distance xt, which is 0.20 m larger than the distance xh traveled
by the hare. Hence,
xt = xh + 0.20 m
which becomes
or
vtt = vh(t โ 120 s) + 0.20 m
(0.100 m/s)t = (2.0 m/s)(t โ 120 s) + 0.20 m
This gives the time of the race as t = 1.30 ร 102 s
(b) xt = vtt = (0.100 m/s)(1.3 ร 102 s) = 13 m
2.15
The maximum allowed time to complete the trip is
t total =
total distance
1600 m โ 1 km/h โ
=
= 23.0 s
required averagespeed 250 km/h โโ 0.278 m/s โโ
The time spent in the first half of the trip is
t1 =
half distance
800 m โ 1 km/h โ
=
= 12.5 s
v1
230 km/h โโ 0.278 m/s โโ
Thus, the maximum time that can be spent on the second half of the trip
is
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Topic 2
55
t2 = ttotal โ t1 = 23.0 s โ 12.5 s = 10.5 s
and the required average speed on the second half is
v2 =
2.16
โ 1 km/h โ
half distance 800 m
=
= 76.2 m/s โ
= 274 km/h
t2
10.5 s
โ 0.278 m/s โโ
(a) From the first kinematic equation with v0 = 0, t = 7.00 ร 10-7 s, and a =
5.30 ร 107 m/s2, the maximum speed is
v = v0 + at = (5.30 ร 107 m/s2 )(7.00 ร 10โ7 s)
v = 37.1 m/s
(b) Use ฮx = v0t + 12 at with v0 = 0 to find the distance travelled during
2
the acceleration:
(
)(
ฮx = 12 at 2 = 12 5.30 ร 107 m/s2 7.00 ร 10โ7 s
)
2
ฮx = 1.30 ร 10โ5 m
2.17
The instantaneous velocity at any time is the slope of the x vs. t graph at
that time. We compute this slope by using two points on a straight
segment of the curve, one point on each side of the point of interest.
(a)
vt=1.00 s =
10.0 m โ 0
= 5.00 m/s
2.00 s โ 0
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Topic 2
56
(b) vt=3.00 s =
(c)
vt=4.50 s =
(d) vt=7.50 s =
2.18
(5.00 โ 10.0) m
= โ2.50 m/s
(4.00 โ 2.00) s
(5.00 โ 5.00) m
= 0
(5.00 โ 4.00) s
0 โ (โ5.00) m
= 5.00 m/s
(8.00 โ 7.00) s
(a) A few typical values are
t (s)
x (m)
1.00
5.75
2.00
16.0
3.00
35.3
4.00
68.0
5.00
119
6.00
192
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publicly accessible website, in whole or in part.
Topic 2
57
(b) We will use a 0.400 s interval centered at t = 400 s. We find at t =
3.80 s, x = 60.2 m and at t = 4.20 s, x = 76.6 m. Therefore,
v=
ฮx 16.4 m
=
= 41.0 m/s
ฮt 0.400 s
Using a time interval of 0.200 s, we find the corresponding values to
be: at t = 3.90 s, x = 64.0 m and at t = 4.10 s, x = 72.2 m. Thus,
v=
ฮx 8.20 m
=
= 41.0 m/s
ฮt 0.200 s
For a time interval of 0.100 s, the values are: at t = 3.95 s, x = 66.0 m,
and at t = 4.05 s, x = 70.1 m. Therefore,
v=
ฮx 4.10 m
=
= 41.0 m/s
ฮt 0.100 s
(c) At t = 4.00 s, x = 68.0 m. Thus, for the first 4.00 s,
v=
ฮx 68.0 m โ 0
=
= 17.0 m/s
ฮt
4.00 s โ 0
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Topic 2
58
This value is much less than the instantaneous velocity at t = 4.00 s.
2.19
Choose a coordinate axis with the origin at the flagpole and east as the
1 2
positive direction. Then, using x = x0 + v0t at with a = 0 for each runner,
2
the x-coordinate of each runner at time t is
xA = โ4.0 mi + (6.0 mi/h)t and xB = 3.0 mi + (โ5.0 mi/h)t
When the runners meet, xA = xB giving
โ4.0 mi + (60 mi/h)t = 3.0 mi + (โ5.0 mi/h)t
or
(6.0 mi/h + 5.0 mi/h)t = 3.0 mi + 4.0 mi. This gives the elapsed time
when they meet as t = (7.0 mi)/(11.0 mi/h) = 0.64 h. At this time,
xA = xB = โ0.18 mi. Thus, they meet 0.18 mi west of the flagpole.
2.20
From the figure below, observe that the motion of this particle can be
broken into three distinct time intervals, during each of which the particle
has a constant acceleration. These intervals and the associated
accelerations are
0 โค t โค 10.0 s, a = a1 = +2.00 m/s2
10 โค t โค 15.0 s, a = a2 = 0
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Topic 2
59
and
15.0 โค t โค 20.0 s, a = a3 = โ3.00 m/s2
(a) Applying vf = vi + a(ฮt) to each of the three time intervals gives
for 0 โค t โค 10.0 s,
v10 + v0 + a1(ฮt1) = 0 + (2.00 m/s2)(10.0 s) = 20.0 m/s
for 10.0 s โค t โค 15.0 s,
v15 + v10 + a2(ฮt2) = 20.0 m/s + 0 = 20.0 m/s
for 15.0 s โค t โค 20.0 s,
v20 + v15 + a3(ฮt3) = 20.0 m/s + (โ3.00 m/s2)(5.00 s) = 5.00 m/s
1
2
(b) Applying ฮx = vi (ฮt) + a(ฮt) to each of the time intervals gives
2
for 0 โค t โค 10.0 s,
1
1
ฮx1 = v0 ฮt1 + a1 (ฮt1 )2 = 0 + (2.00 m/s 2 )(10.0 s)2 = 1.00 ร 10 2 m
2
2
for 10.0 s โค t โค 15.0 s,
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Topic 2
60
1
ฮx2 = v10 ฮt 2 + a2 (ฮt 2 )2 = (20.0 m/s)(5.00 s) + 0 = 1.00 ร 10 2 m
2
for 15.0 s โค t โค 20.0 s,
1
ฮx = v15 ฮt 3 + a3 (ฮt 3 )2
2
1
= (20.0 m/s)(5.00 s) + (โ3.00 m/s 2 )(5.00 s)2 = 62.5 m
2
Thus, the total distance traveled in the first 20.0 s is
ฮxtotal = ฮx1 + ฮx2 + ฮx3 = 100 m + 100 m + 62.5 m = 263 m
2.21
We choose the positive direction to point away from the wall. Then, the
initial velocity of the ball is vi = โ25.0 m/s and the final velocity is vf =
+22.0 m/s. If this change in velocity occurs over a time interval of ฮt = 3.50
ms (i.e., the interval during which the ball is in contact with the wall), the
average acceleration is
a=
2.22
ฮv v f โ vi +22.0 m/s โ (โ25.0 m/s)
=
=
= 1.34 ร 10 4 m/s 2
โ3
ฮt
ฮt
3.50 ร 10 s
From a = ฮv/ฮt, the required time is
ฮt =
2.23
From a =
โ โ 0.447 m/s โ
ฮv โ 60.0 mi/h โ 0 โ โ
1g
=โ
=
= 0.391 s
2โ
โ
โ
โ โ 9.80 m/s โ โโ 1 mi/h โโ
a โ
7g
ฮv
ฮv (60 โ 55) mi/h โ 0.447 m/s โ
=
= 3.7 s
, we have ฮt =
a
0.60 m/s 2 โโ 1 mi/h โโ
ฮt
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publicly accessible website, in whole or in part.
Topic 2
2.24
61
(i)
(a)
From t = 0 to t = 5.0 s,
a=
v f โ vi โ8.0 m/s โ (โ8.0 m/s)
=
= 0
t f โ ti
5.0 s โ 0
(b) From t = 5.0 s to t = 15 s,
a=
8.0 m/s โ (โ8.0 m/s)
= 1.6 m/s 2
15 s โ 5.0 s
(c) From t = 0 to t = 20 s,
a=
8.0 m/s โ (โ8.0 m/s)
= 0.80 m/s 2
20 s โ 0
(ii) At any instant, the instantaneous acceleration equals the slope of the
line tangent to the v vs. t graph at that point in time.
(a) At t = 2.0 s, the slope of the tangent line to the curve is 0.
(b) At t = 10 s, the slope of the tangent line is 1.6 m/s2 .
(c) At t = 18 s, the slope of the tangent line is 0.
2.25
(a) a =
ฮv 175 mi/h โ 0
=
= 70.0 mi/h โ
s
ฮt
2.5 s
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Topic 2
62
โ
mi โ โ 1609 m โ โ 1 h โ
a = โ 70.0
=
= 31.3 m/s 2
h โ
s โโ โโ 1 mi โโ โโ 3600 s โโ
โ
or
Alternatively,
mโ
1g
โ
a = โ 31.3 2 โ =
= 3.19g
โ
s โ 9.80 m/s 2
1 2
(b) If the acceleration is constant, ฮx = v0t + at
2
1โ
mโ
ฮx = 0 + โ 3.13 2 โ (2.50 s)2 = 97.8 m
2โ
s โ
or
2.26
โ 3.281 ft โ
ฮx = (97.8 m) โ
= 321 ft
โ 1 m โโ
As in the algebraic solution to Example 2.5, we let t represent the time the
trooper has been moving.
We graph
xcar = 24.0 m + (24.0 m/s)t
and
xtrooper = (1.50 m/s2)t2
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Topic 2
63
The curves intersect at t = 16.9 s
2.27
1 2
Apply ฮx = v0 + at to the 2.00-second time interval during which the
2
object moves from xi = 3.00 cm to xf = โ5.00 cm. With v0 = 12.0 cm/s, this
yields an acceleration of
2 โก(x f โ xi ) โ v0t โคโฆ 2[(โ5.00 โ 3.00) cm โ (12.0 cm/s)(2.00 s)]
a= โฃ
=
t2
(2.00 s)2
or
2.28
a = โ16.0 cm/s2
From v 2 = v02 + 2a(ฮx) , we have (10.97 ร 103 m/s)2 = 0 + 2a(220 m) so that
a=
v 2 โ v02 (10.97 ร 10 3 m/s)2 โ 0
=
= 2.74 ร 10 5 m/s 2
2(ฮx)
2(220 m)
โ
โ
1g
= (2.74 ร 10 5 m/s 2 ) โ
= 2.79 ร 10 4 times g!
2โ
โ 9.80 m/s โ
2.29
(a)
ฮx = v(ฮt) = [(v + v0 ) / 2]ฮt becomes
โ 2.80 m/s + v0 โ
40.0 m= โ
โโ (8.50 s)
โ
2
which yields
(b) a =
v0 =
2
(40.0 m) โ 2.80 m/s = 6.61 m/s
8.50 s
v โ v0 2.80 m/s โ 6.61 m/s
=
= โ0.448 m/s 2
ฮt
8.50 s
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Topic 2
2.30
64
(a)
(b) The known quantities are initial velocity, final velocity, and
displacement. The kinematics equation that relates these quantities to
acceleration is v f = vi + 2a(ฮx)
2
(c)
2
v 2f โ vi2
a=
2(ฮx)
(d) a =
v 2f โ vi2 (30.0 m/s)2 โ (20.0 m/s)2
=
= 1.25 m/s 2
2(ฮx)
2(2.00 ร 10 2 m)
(e) Using a = ฮv/ฮt, we find that
ฮt =
2.31
ฮv v f โ vi 30.0 m/s โ 20.0 m/s
=
=
= 8.00 s
a
a
1.25 m/s 2
(a) With v = 120 km/h, v 2 = v02 + 2a(ฮx) yields
v 2 โ v02 โกโฃ(120 km/h) โ 0 โคโฆ โ 0.278 m/s โ
a=
=
= 2.32 m/s 2
โ
โ
2(ฮx)
2(240 m)
โ 1 km/h โ
2
2
(b) The required time is
ฮt =
2.32
v โ v0 (120 km/h โ 0) โ 0.278 m/s โ
=
= 14.4 s
a
2.32 m/s 2 โโ 1 km/h โโ
(a) From v f = vi + 2a(ฮx) , with vi = 6.00 m/s and vf = 12.0 m/s, we find
2
2
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Topic 2
65
ฮx =
v 2f โ vi2 (120 m/s)2 โ (6.00 m/s)2
=
= 13.5 m
2a
2(4.00 m/s 2 )
(b) In this case, the object moves in the same direction for the entire time
interval and the total distance traveled is simply the magnitude or
absolute value of the displacement. That is,
d = |ฮx| = 13.5 m
(c) Here, vi = โ6.00 m/s and vf = 12.0 m/s, and we find
v 2f โ vi2
ฮx =
= 13.5m
2a
[the same as in part (a)]
(d) In this case, the object initially slows down as it travels in the
negative x-direction, stops momentarily, and then gains speed as it
begins traveling in the positive x-direction. We find the total distance
traveled by first finding the displacement during each phase of this
motion.
While coming to rest (vi = โ6.00 m/s, vf = 0),
v 2f โ vi2 (0)2 โ (โ6.00 m/s)2
ฮx1 =
=
= โ4.50 m
2a
2(4.00 m/s2 )
After reversing direction (vi = 0 m/s, vf = 12.0 m/s),
v 2f โ vi2 (12.0 m/s)2 โ (0)2
ฮx2 =
=
= 18.0 m
2a
2(4.00 m/s 2 )
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Topic 2
66
Note that the net displacement is ฮx = ฮx1 + ฮx2 = โ4.50 m + 18.0 m =
13.5 m, as found in part (c) above. However, the total distance
traveled in this case is
d = |ฮx1| + |ฮx2| = |โ4.50 m| + |18.0 m| = 22.5 m
2.33
v โ v0 24.0 m/s 2 โ 0
=
= 8.14 m / s 2
(a) a =
ฮt
2.95 s
(b) From a = ฮv/ฮt, the required time is
ฮt =
v f โ vi 20.0 m/s โ 10.0 m/s
=
= 1.23 s
a
8.15 m/s 2
(c) Yes. For uniform acceleration, the change in velocity ฮv generated in
time ฮt is given by ฮv = a(ฮt). From this, it is seen that doubling the
length of the time interval ฮt will always double the change in
velocity ฮv. A more precise way of stating this is: โWhen acceleration
is constant, velocity is a linear function of time.โ
2.34
(a) The time required to stop the plane is
t=
v โ v0 0 โ 100 m/s
=
= 20.0 s
a
โ5.00 m/s 2
(b) The minimum distance needed to stop is
โ v + v0 โ
โ 0 + 100 m/s โ
ฮx = v = โ
t=โ
โ
โโ (20.0 s) = 1 000 m = 1.00 km
โ 2 โ
โ
2
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Topic 2
67
Thus, the plane requires a minimum runway length of 1.00 km.
It cannot land safely on a 0.800 km runway.
2.35
We choose x = 0 and t = 0 at the location of Sueโs car when she first spots
the van and applies the brakes. Then, the initial conditions for Sueโs car
are x0S = 0 and v0S = 30.0 m/s. Her constant acceleration for aS = โ2.00 m/s2.
The initial conditions for the van are x0V = 155 m, v0V = 5.00 m/s, and its
1 2
constant acceleration is aV = 0. We then use ฮx = x โ x0 = v0t + at to write
2
an equation for the x-coordinate of each vehicle for t โฅ 0. This gives
Sueโs Car:
1
xS โ 0 = (30.0 m/s)t + (โ2.00 m/s 2 )t 2 or xS = (30.0 m/s)t โ (โ1.00 m/s 2 )t 2
2
xV โ 155 m = (5.00 m/s)t +
Van:
1
(0)t 2 or xV = 155 m + (5.00 m/s)t
2
In order for a collision to occur, the two vehicles must be at the same
location (i.e., xS = xV). Thus, we test for a collision by equating the two
equations for the x-coordinates and see if the resulting equation has any
real solutions.
xS = xV
โ
(30.0 m/s)t โ (1.00 m/s2)t2 = 155 m + (5.00 m/s)t
or
(1.00 m/s2)t2 โ (25.00 m/s) + 155 m = 0
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Topic 2
68
Using the quadratic formula yields
โ(โ25.00 m/s) ยฑ (โ25.00 m/s 2 ) โ 4(1.00 m/s 2 )(155 m)
t=
= 13.6s or 11.4 s
2(1.00 m/s 2 )
The solutions are real, not imaginary, so a collision will occur. The
smaller of the two solutions is the collision time. (The larger solution tells
when the van would pull ahead of the car again if the vehicles could pass
harmlessly through each other.) The x-coordinate where the collision
occurs is given by
xcollision = xS|t= 11.4 s = xV|t= 11.4 s = 155 m + (5.00 m/s)(11.4 s) = 212 m
2.36
The velocity at the end of the first interval is
v = v0 + at = 0 + (2.77 m/s2)(15.0 s) = 41.6 m/s
This is also the constant velocity during the second interval and the initial
velocity for the third interval. Also, note that the duration of the second
interval is t2 = (2.05 min)(60.0 s/1 min) = 123 s.
1 2
(a) From ฮx = v0t + at , the total displacement is
2
(ฮx)total = (ฮx)1 + (ฮx)2 + (ฮx)3
1
โก
โค
โ โข 0 + (2.77 m/s 2 )(15.0 s)2 โฅ + [(2.77 m/s 2 )(15.0 s)(123 s) + 0]
2
โฃ
โฆ
1
โก
โค
+ โข(2.77 m/s 2 )(15.0 s)(4.39 s) + (โ9.47 m/s 2 )(4.39 s)2 โฅ
2
โฃ
โฆ
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Topic 2
69
(ฮx)total = 312 m + 5.11 ร 103 m + 91.2 m = 5.51 ร 103 m = 5.51 km
or
(b) v1 =
(ฮx)1 312 m
=
= 20.8 m/s
t1
15.0 s
v2 =
(ฮx)2 5.11ร 10 3 m
=
= 41.5 m/s
t2
123 s
v3 =
(ฮx)3 91.2 m
=
= 20.8 m/s , and the average velocity for the
t3
4.39 s
(ฮx)total
5.51ร 10 3 m
=
= 38.7 m/s
total trip is vtotal =
t total
(15.0 + 123 + 4.39) s
2.37
1 2
Using the uniformly accelerated motion equation ฮx + v0t + at for the
2
1
2
2
full 40 s interval yields ฮx = (20 m/s)(40 s) + (โ1.0 m/s )(40 s) = 0 , which
2
is obviously wrong. The source of the error is found by computing the
time required for the train to come to rest. This time is
t=
v โ v0 0 โ 20 m/s
=
= 20 s.
a
โ1.0 m/s 2
Thus, the train is slowing down for the first 20 s and is at rest for the last
20 s of the 40 s interval.
The acceleration is not constant during the full 40 s. It is, however,
constant during the first 20 s as the train slows to rest. Application of
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Topic 2
70
1
ฮx = v0t + at 2 to this interval gives the stopping distance as
2
1
ฮx = (20 m/s)(20 s) + (โ1.0 m/s 2 )(20 s)2 = 200 m
2
2.38
mi โ โ 0.447 m/s โ
โ
v0 = 0 and v f = โ 400 โ โ
= 17.9 m/s
โ
h โ โ 1 mi/h โโ
(a) To find the distance traveled, we use
โ v + v0 โ โ 17.9 m/s + 0 โ
ฮx = vt = โ f
t=โ
โโ (12.0 s) = 107 m
โ 2 โโ โ
2
(b) The constant acceleration is
2.39
a=
v f โ v0 17.9 m/s โ 0
=
= 1.49 m/s 2
t
12.0 s
At the end of the acceleration period, the velocity is
v = v0 + ataccel = 0 + (1.5 m/s2)(5.0 s) = 7.5 m/s
This is also the initial velocity for the braking period.
(a) After braking,
vf = v + atbrake = 7.5 m/s + (โ2.0 m/s2)(3.0 s) = 1.5 m/s
(b) The total distance traveled is
โ v + vโ
โ v + v0 โ
ฮxtotal = (ฮx)accel + (ฮx)brake = (vt)accel + (vt)brake = โ
t accel + โ f
t
โ
โ 2 โ
โ 2 โโ brake
โ 7.5 m/s +0 โ
โ 1.5 m/s +7.5 m/s โ
ฮx total = โ
(5.0 s) + โ
โ
โโ (3.0 s) = 32 m
โ
โ
โ
2
2
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Topic 2
2.40
71
For the acceleration period, the parameters for the car are: initial velocity
= via = 0, acceleration = aa = a1, elapsed time = (ฮt)a = t1, and final velocity =
vfa. For the braking period, the parameters are: initial velocity = vib = final
velocity of acceleration period = vfa, acceleration = ab = a2, and elapsed
time = (ฮt)b = t2.
(a) To determine the velocity of the car just before the brakes are
engaged, we apply vf = vi + a(ฮt) to the acceleration period and find
vib = vfa = via + aa(ฮt)a = 0 + a1t1
or
vib = a1t1
1
2
(b) We may use ฮx = vi (ฮt) + a(ฮt) to determine the distance traveled
2
during the acceleration period (i.e., before the driver begins to brake).
This gives
1
1
(ฮx)a = via (ฮt)a + aa (ฮt)2a = 0 + a1t12
2
2
or
(ฮx)a =
1 2
a1t1
2
(c) The displacement occurring during the braking period is
1
1
(ฮx)b = vib (ฮt)b + ab (ฮt)b2 = (a1t1 )t 2 + a2t 22
2
2
Thus, the total displacement of the car during the two intervals
combined is
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Topic 2
72
(ฮx)total = (ฮx)a + (ฮx)b =
2.41
1 2
1
a1t1 + a1t1t 2 + a2t 22
2
2
The time the Thunderbird spends slowing down is
ฮt1 =
ฮx1 2(ฮx1 )
2(250 m)
=
=
= 6.99 s
v1
v + v0 0 + 71.5 m/s
The time required to regain speed after the pit stop is
ฮt 2 =
ฮx2 2(ฮx2 )
2(350 m)
=
=
= 9.79 s.
v2
v + v0 71.5 m/s + 0
Thus, the total elapsed time before the Thunderbird is back up to speed is
ฮt = ฮt1 + 5.00 s + ฮt2 = 6.99 s + 5.00 s + 9.79 s = 2.18 s
During this time, the Mercedes has traveled (at constant speed) a distance
ฮxM = v0(ฮt) = (71.5 m/s)(21.8 s) = 1 559 m
and the Thunderbird has fallen behind a distance
d = ฮxM โ ฮx1 โ ฮx2 = 1 559 m โ 250 m โ 350 m = 959 m
2.42
The initial bank account balance is xi = $1.0 ร 104 (to two significant
figures) and the bank account is empty when xf = 0 with a change of ฮx =
xf โ xi = $(โ1.0 ร 104). Use
ฮx = v0t + 12 at 2
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Topic 2
73
with v0 = 0 and a = โ2.5 ร 102 $/month to find the time, t:
ฮx = 12 at 2 โ t =
2ฮx
a
(
)
t=
2.43
2 $(โ1.0 ร 10 4 )
( โ2.5 ร 10 $/month )
2
2
= 8.9 months
(a) Take t = 0 at the time when the player starts to chase his opponent. At
this time, the opponent is distance d = (12 m/s)(30 s) = 36 m in front of
the player. At time t > 0, the displacements of the players from their
initial positions are
1
1
ฮxplayer = (v0 )player t + aplayer t 2 = 0 + (4.0 m/s 2 )t 2
2
2
1
ฮxopponent = (v0 )opponent t + aopponent t 2 = (12 m/s)t + 0
2
and
When the players are side-by-side, ฮxplayer = ฮxopponent + 36 m
[1]
[2]
[3]
Substituting Equations [1] and [2] into Equation [3] gives
1
(4.0 m/s 2 )t 2 = (12 m/s)t + 36 m
2
or
t2 + (โ6.0 s)t + (โ18 s2) = 0
Applying the quadratic formula to this result gives
t=
โ(โ6.0 s) ยฑ (โ6.0 s)2 โ 4(1)(โ18 s 2 )
2(1)
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Topic 2
74
which has solutions of t = โ2.2 s and t = +8.2 s. Since the time must be
greater than zero, we must choose t = 8.2 s as the proper answer.
1
1
2
2
2
2
(b) ฮxplayer =(v0 )player t + aplayer t = 0 + (4.0 m/s )(8.2 s) = 1.3 ร 10 m
2
2
2.44
The initial velocity of the train is v0 = 82.4 and the final velocity is v = 16.4
km/h. The time required for the 400 m train to pass the crossing is found
from ฮx = vt = [(v + v0 )/2]t as
t=
2.45
โ 3600 s โ
2(ฮx)
2(0.400 km)
=
= (8.10 ร 10 โ3 h) โ
= 29.1 s
v + v0 (82.4 + 16.4) km/h
โ 1 h โโ
(a) From v 2 = v02 + 2a(ฮy) with v = 0, we have
v 2 โ v02 0 โ (25.0 m/s)2
(ฮy)max =
=
= 31.9 m
2a
2(โ9.80 m/s 2 )
(b) The time to reach the highest point is
v 2 โ v02 0 โ 25.0 m/s
t up =
=
= 2.55 s
a
โ9.80 m/s 2
(c) The time required for the ball to fall 31.9 m, starting from rest, is
found from
1
ฮy = (0)t + at 2 as t =
2
2(ฮy)
=
a
2(โ31.9 m)
= 2.55 s
โ9.80 m/s 2
(d) The velocity of the ball when it returns to the original level (2.55 s
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Topic 2
75
after it starts to fall from rest) is
v = v0 + at = 0 + (โ9.80 m/s2)(2.55 s) = โ25.0 m/s
2.46
We take upward as the positive y-direction and y = 0 at the point where
the ball is released. Then, v0y = โ8.00 m/s, ay = โg = โ9.80 m/s2, and ฮy =
โ30.0 m when the ball reaches the ground. From v y = v0 y + 2ay (ฮy) , the
2
2
velocity of the ball just before it hits the ground is
v y = v02 y + 2ay (ฮy) = โ (8.00 m/s)2 + 2(โ9.80 m/s 2 )(โ30.0 m) = โ25.5 m/s
Then, vy = v0y + ayt gives the elapsed time as
t=
2.47
v y โ v0 y โ25.5 m/s โ (โ8.00 m/s)
=
= 1.79 s
ay
โ9.80 m/s 2
(a) The velocity of the object when it was 30.0 m above the ground can be
1 2
determined by applying ฮy = v0t + at to the last 1.50 s of the fall.
2
This gives
1โ
mโ
โ30.0 m = v0 (1.50 s) + โ โ9.80 2 โ (1.50 s)2
2โ
s โ
or
v0 = โ12.7 m/s
(b) The displacement the object must have undergone, starting from rest,
to achieve this velocity at a point 30.0 m above the ground is given by
v 2 = v02 + 2a(ฮy) as
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Topic 2
76
(ฮy)1 =
v 2 โ v02 (โ12.7 m/s)2 โ 0
=
= โ8.23 m
2a
2(โ9.80 m/s 2 )
The total distance the object drops during the fall is then
|(ฮy)total|=|(โ8.23 m) + (โ30.0 m)| = 38.2 m
2.48
(a) Consider the rockโs entire upward flight, for which v0 = +7.40 m/s,
vf = 0, a = โg = โ9.80 m/s2, yi = 1.55 m, (taking y = 0 at ground level),
and yf = hmax = maximum altitude reached. Then applying
v 2f = vi2 + 2a(ฮy) to this upward flight gives
0 = (7.40 m/s)2 + 2(โ9.80 m/s2)(hmax โ 1.55 m)
Solving for the maximum altitude of the rock gives
hmax = 1.55 m +
(7.40 m/s)2
= 4.34 m
2(9.80 m/s 2 )
Since hmax > 3.65 m (height of the wall), the rock does reach the top of
the wall.
(b) To find the velocity of the rock when it reaches the top of the wall, we
use v f = vi + 2a(ฮy) and solve for vf when yf = 3.65 m (starting with vi
2
2
= +7.40 m/s at yi = 1.55 m). This yields
v f = vi2 + 2a(y f โ yi ) = (7.40 m/s)2 + 2(โ9.80 m/s 2 )(3.65 m โ 1.55 m) = 3.69 m/s
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Topic 2
77
(c) A rock thrown downward at a speed of 7.40 m/s (vi = โ7.40 m/s) from
the top of the wall undergoes a displacement of (ฮy) = yf โ yi =
1.55 m โ 3.65 m = โ2.10 m before reaching the level of the attacker. Its
velocity when it reaches the attacker is
v f = โ vi2 + 2a(ฮy) = โ (โ7.40 m/s)2 + 2(โ9.80 m/s 2 )(โ2.10 m) = โ9.79 m/s
so the change in speed of this rock as it goes between the 2 points
located at the top of the wall and the attacker is given by
ฮ(speed)down = ||vf| โ |vi|| = ||โ9.79 m/s| โ |โ7.40 m/s|| = 2.39 m/s.
(d) Observe that the change in speed of the ball thrown upward as it
went from the attacker to the top of the wall was
ฮ(speed)up = ||vf| โ |vi||=|3.69 m/s โ 7.40 m/s| = 3.71 m/s
The two rocks do not undergo the same magnitude speed change.
The rocks have the same acceleration, but the rock thrown downward
has a higher average speed between the two levels, and is accelerated
over a smaller time interval.
2.49
The velocity of the childโs head just before impact (after falling a distance
of 0.40 m, starting from rest) is given by v 2 = v02 + 2a(ฮy) as
v I = v02 + 2a(ฮy) = โ 0 + 2(โ9.8 m/s 2 )(โ0.40 m) = โ2.8 m/s
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Topic 2
78
If, upon impact, the childโs head undergoes an additional displacement
ฮy = โh before coming to rest, the acceleration during the impact can be
found from v 2 = v02 + 2a(ฮy) to be a = (0 โ v I2 )/2(โh) = v I2 /2h . The duration
of the impact is found from v = v0 + at as
t = ฮv/a = โv I /(v I2 /2h), or t = โ2h/v I .
Applying these results to the two cases yields
Hardwood Floor
(h = 2.0 ร 10 โ3 m) : a =
and
t=
vI2
(โ2.8 m/s)2
=
= 2.0 ร 10 3 m/s 2
2h 2(2.0 ร 10 โ3 m)
โ2h โ2(2.0 ร 10 โ3 m)
=
= 1.4 ร 10 โ3 s = 1.4 ms
vI
โ2.8 m/s
Carpeted Floor (h = 1.0 ร 10
and
2.50
t=
โ2
m): a =
vI2
(โ2.8 m/s)2
=
= 3.9 ร 10 2 m/s 2
โ2
2h 2(1.0 ร 10 m)
โ2h โ2(1.0 ร 10 โ2 m)
=
= 7.1ร 10 โ3 s = 7.1 ms
vI
โ2.8 m/s
(a) After 2.00 s, the velocity of the mailbag is
vbag = v0 + at = โ1.50 m/s + (โ9.80 m/s2)(2.00 s) = โ21.1 m/s
The negative sign tells us that the bag is moving downward and the
magnitude of the velocity gives the speed as 21.1 m/s.
(b) The displacement of the mailbag after 2.00 s is
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Topic 2
79
โ v + v0 โ
โก โ21.1 m/s + ( โ 1.50 m/s) โค
(ฮy)bag = โ
t=โข
โ
โฅโฆ (2.00 s) = โ 22.6 m
โ 2 โ
2
โฃ
During this time, the helicopter, moving downward with constant
velocity, undergoes a displacement of
1
(ฮy)copter = v0t + at 2 = (โ1.5 m/s)(2.00 s) + 0 = โ3.00 m
2
The distance separating the package and the helicopter at this time is
then
d = |(ฮy)P โ (ฮy)h| = |โ22.6 m โ (โ3.00 m)| = |โ19.6 m| = 19.6 m
(c) Here, (v0)bag = (v0)copter = +1.50 m/s and abag = โ9.80 m/s2 while acopter = 0.
After 2.00 s, the velocity of the mailbag is
vbag = 1.50
m โ
mโ
m
m
+ โ โ9.80 2 โ (2.00 s) = โ18.1 and its speed is vbag = 18.1
s โ
s โ
s
s
In this case, the displacement of the helicopter during the 2.00 s
interval is
ฮycopier = (+1.50 m/s)(2.00 s) + 0 = +3.00 m
Meanwhile, the mailbag has a displacement of
โ v + v0 โ
โก โ18.1 m/s + 1.50 m/s โค
(ฮy)bag = โ bag
t=โข
โ
โฅโฆ (2.00 s) = โ16.6 m
โ
2 โ
2
โฃ
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Topic 2
80
The distance separating the package and the helicopter at this time is
then
d = |(ฮy)p โ (ฮy)h| = |โ16.6 m โ (+3.00 m)| = |โ19.6 m| = 19.6 m
2.51
(a) From the instant the ball leaves the playerโs hand until it is caught,
the ball is a freely falling body with an acceleration of
a = โg = โ9.80 m/s2 = 9.80 m/s2 = 9.80 m/s2 (downward)
(b) At its maximum height, the ball comes to rest momentarily and then
= 0 .
begins to fall back downward. Thus, vmax
height
1 2
(c) Consider the relation ฮy = v0t + at with a = โg. When the ball is at
2
the throwerโs hand, the displacement is ฮy = 0, giving
1
0 = v0t โ gt 2
2
This equation has two solutions, t = 0, which corresponds to when the
ball was thrown, and t = 2v0/g corresponding to when the ball is
caught. Therefore, if the ball is caught at t = 2.00 s, the initial velocity
must have been
v0 =
gt (9.80 m/s 2 )(2.00 s)
=
= 9.80 m/s
2
2
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Topic 2
81
(d) From v 2 = v02 + 2a(ฮy) , with v = 0 at the maximum height,
(ฮy)max =
2.52
v 2 โ v02 0 โ (9.80 m/s 2 )
=
= 4.90 m
2a
2(โ9.80 m/s 2 )
(a) Let t = 0 be the instant the package leaves the helicopter, so the
package and the helicopter have a common initial velocity of vi = โv0
(choosing upward as positive).
At times t > 0, the velocity of the package (in free-fall with constant
acceleration ap = โg) is given by v = vi + at as vp = โv0 โ gt = โ(v0 + gt)
and speed = |vp| = v0 + gt.
(b) After an elapsed time t, the downward displacement of the package
from its point of release will be
1
1
1 โ
โ
(ฮy)P = vi t + aP t 2 = โv0t โ gt 2 = โ โ v0t + gt 2 โ
โ
2
2
2 โ
and the downward displacement of the helicopter (moving with
constant velocity, or acceleration ah = 0) from the release point at this
time is
1
(ฮy)h = vi t + aht 2 = โv0t + 0 = โv0t
2
The distance separating the package and the helicopter at this time is
then
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Topic 2
82
1
1
โ
โ
d = (ฮy) p โ (ฮy)h = โ โ v0t + gt 2 โ โ (โv0t) = gt 2
โ
โ
2
2
(c) If the helicopter and package are moving upward at the instant of
release, then the common initial velocity is vi = +v0. The accelerations
of the helicopter (moving with constant velocity) and the package (a
freely falling object) remain unchanged from the previous case (ap =
โg and ah = 0).
In this case, the package speed at time t > 0 is |vP| = |vi + apt| =
|v0 โ gt| .
At this time, the displacements from the release point of the package
and the helicopter are given by
1
1
(ฮy) p = vi t + a pt 2 = v0t โ gt 2
2
2
and
1
(ฮy)h = vi t + aht 2 = v0t + 0 = +v0t
2
The distance separating the package and helicopter at time t is now
given by
1
1
d = (ฮy) p โ (ฮy)h = v0t โ gt 2 โ v0t = gt 2
2
2
2.53
(the same as earlier!)
(a) After its engines stop, the rocket is a freely falling body. It continues
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Topic 2
83
upward, slowing under the influence of gravity until it comes to rest
momentarily at its maximum altitude. Then it falls back to Earth,
gaining speed as it falls.
(b) When it reaches a height of 150 m, the speed of the rocket is
v = v02 + 2a(ฮy) = (50.0 m/s)2 + 2(2.00 m/s 2 )(150 m) = 55.7 m/s
After the engines stop, the rocket continues moving upward with an
initial velocity of v0 = 55.7 m/s and acceleration a = โg = โ9.80 m/s2.
When the rocket reaches maximum height, v = 0. The displacement of
the rocket above the point where the engines stopped (that is, above
the 150 m level) is
ฮy =
v 2 โ v02 0 โ (55.7 m/s)2
=
= 158 m
2a
2(โ9.80 m/s 2 )
The maximum height above ground that the rocket reaches is then
given by hmax = 150 m + 158 m = 308 m
(c) The total time of the upward motion of the rocket is the sum of two
intervals. The first is the time for the rocket to go from v0 = 50.0 m/s at
the ground to a velocity of v = 55.7 m/s at an altitude of 150 m. This
time is given by
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Topic 2
84
t1 =
(ฮy)1
(ฮy)1
2(150 m)
=
=
= 2.84 s
v1
(v + v0 ) / 2 (55.7 + 50.0) m/s
The second interval is the time to rise 158 m starting with v0 = 55.7
m/s and ending with v = 0. This time is
t2 =
(ฮy)2
(ฮy)2
2(158 m)
=
=
= 5.67 s
v2
(v + v0 ) / 2 0 + 55.7 m/s
The total time of the upward flight is then tap = t1 + t2 =
(2.84 + 5.67) s = 8.51 s
(d) The time for the rocket to fall 308 m back to the ground, with v0 = 0
1 2
and acceleration a = โg = โ9.80 m/s2, is found from ฮy = v0t + at as
2
t down =
2(ฮy)
=
a
2(โ308 m)
= 7.93 s
โ9.80 m/s 2
so the total time of the flight is tflight = tup + tdown = (8.51 + 7.93) s =
16.4 s.
2.54
(a) For the upward flight of the ball, we have vi = v0, vf = 0, a = โg, and ฮt
= 3.00 s. Thus, vf = vi + a(ฮt) gives the initial velocity as
vi = vf โ a(ฮt) = vf + g(ฮt) or v0 = 0 + (9.80 m/s2)(3.00 s) = +29.4 m/s
(b) The vertical displacement of the ball during this 3.00-s upward flight
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Topic 2
85
is
โ v + vf โ
โ 29.4 m/s + 0 โ
ฮymax = h = v(ฮt) = โ i
(ฮt) = โ
โโ (3.00 s) = 44.1 m
โ
โ
โ 2 โ
2
2.55
During the 0.600 s required for the rig to pass completely onto the bridge,
the front bumper of the tractor moves a distance equal to the length of the
rig at constant velocity of v = 100 km/h. Therefore the length of the rig is
โก
km โ 0.278 m/s โ โค
Lrig = vt = โข100
โโ 1 km/h โโ โฅ (0.600 s) = 16.7 m
h
โฃ
โฆ
While some part of the rig is on the bridge, the front bumper moves a
distance ฮx = Lbridge + Lrig = 100 m + 16.7 m. With a constant velocity of v =
100 km/h, the time for this to occur is
t=
2.56
Lbridge + Lrig 400 m + 16.7 m โ 1 km/h โ
=
โโ 0.278 m/s โโ = 15.0 s
v
100 km/h
(a) The acceleration experienced as he came to rest is given by v = v0 + at
as
a=
v โ v0
=
t
mi โ โ 0.447 m/s โ
โ
0 โ โ 632 โ โ
โ
h โ โ 1 mi/h โโ
1.40 s
= โ202 m/s 2
(b) The distance traveled while stopping is found from
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Topic 2
86
โก โ
mi โ โ 0.447 m/s โ โค
โข 0 + โโ 632 โโ โ
โฅ
h โ 1 mi/h โโ โฆ
โ v + v0 โ
โฃ
ฮx = vt = โ
t=
(1.40 s) = 198 m
โ 2 โโ
2
2.57
(a) The acceleration of the bullet is
a=
v 2 โ v02 (300 m/s)2 โ (400 m/s)2
=
= โ3.50 ร 10 5 m/s 2
2(ฮx)
2(0.100 m)
(b) The time of contact with the board is
t=
2.58
v โ v0 (300 โ 400) m/s
=
= 2.86 ร 10 โ4 s
5
2
a
โ3.50 ร 10 m/s
1 2
(a) From ฮx = v0t + at , we have
2
1
100 m = (30.0 m/s)t + (โ3.50 m/s 2 )t 2
2
This reduces to 3.50t2 + (โ60.0 s)t + (200 s2) = 0, and the quadratic
formula gives
โ(โ60.0 s) ยฑ (โ60.0 s)2 โ 4(3.50)(200 s 2 )
t=
2(3.50)
The desired time is the smaller solution of t = 4.53 s. The larger
solution of t = 12.6 s is the time when the boat would pass the buoy
moving backwards, assuming it maintained a constant acceleration.
(b) The velocity of the boat when it first reaches the buoy is
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Topic 2
87
v = v0 + at = 30.0 m/s + (โ350 m/s2)(4.53 s) = 14.1 m/s.
2.59
(a) The keys have acceleration a = โg = โ9.80 m/s2 from the release point
1 2
until they are caught 1.50 s later. Thus, ฮy = v0t + at gives
2
ฮy โ at 2 /2 (+4.00 m) โ (โ9.80 m/s 2 )(1.50 s)2 /2
v0 =
=
= +10.0 m/s
t
1.50 s
or
v0 = 10.0 m/s upward
(b) The velocity of the keys just before the catch was
v = v0 + at = 10.0 m/s + (โ9.80 m/s2)(1.50 s) = โ4.70 m/s
or
2.60
v = 4.70 m/s downward
(a) While in the air after launching itself from the water, the salmonโs
vertical acceleration is ay = โg = โ9.80 m/s2. Assume it comes to rest at
the top of its vertical leap, a distance ฮy = 3.60 m above the bottom of
the waterfall. From the time-independent kinematic equation, with
the final velocity v = 0, the initial speed v0 is
v 2 = v02 โ 2g ( ฮy )
(
)
v0 = 2g ( ฮy ) = 2 9.80 m/s2 ( 3.6 m )
v0 = 8.4 m/s
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Topic 2
2.61
88
(a) From v 2 = v02 + 2a(ฮy) , the insectโs velocity after straightening its legs
is
v = v02 + 2a(ฮy) = 0 + 2(4 000 m/s 2 )(2.0 ร 10 โ3 m) = 4.0 m/s
(b) The time to reach this velocity is
t=
v โ v0 4.0 m/s โ 0
=
= 1.0 ร 10 โ3 s = 1.0 ms
a
4 000 m/s 2
(c) The upward displacement of the insect between when its feet leave
the ground and it comes to rest momentarily at maximum altitude is
ฮy =
2.62
v 2 โ v02 0 โ v02 โ(4.0 m/s)2
=
=
= 0.82 m
2a
2(โg) 2(โ9.8 m/s 2 )
(a)
For constant speed:
(b)
When speeding up at a constant rate:
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Topic 2
89
(c)
2.63
When slowing down at a constant rate:
The falling ball moves a distance of (15 m โ h) before they meet, where h
1 2
is the height above the ground where they meet. Apply ฮy = v0t + at ,
2
with a = โg, to obtain
1
โ(15 m โ h) = 0 โ gt 2
2
or
1
h =15 m โ gt 2
2
[1]
1 2
Applying ฮy = v0t + at to the rising ball gives
2
1
h = (25 m/s)t โ gt 2
2
[2]
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publicly accessible website, in whole or in part.
Topic 2
90
Combining Equations [1] and [2] gives
(25 m/s)t โ
or
2.64
t=
1 2
1
gt = 15 m โ gt 2
2
2
15 m
= 0.60 s
25 m/s
(a) When the ball hits the ground, its change in height will be ฮy = โh.
Solve for the final speed of each ball using the time-independent
kinematic equation:
v 2 = v02 โ 2g ( ฮy )
v=
( ยฑv ) โ 2g ( โh) =
2
0
v02 + 2gh
(b) Weโre asked to find an expression for the time difference ฮt between
the times of flight for the upward- and downward-thrown balls.
For the upward-thrown ball, the path to the ground can be separated
into two parts. In the first part the ball rises with initial velocity +v0
and falls back to its original height where its velocity is โv0. In the
second part it moves from height h to the ground in exactly the same
time it takes the downward-thrown ball to reach the ground. The
difference between the two ballโs times of flight is therefore equal to
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publicly accessible website, in whole or in part.
Topic 2
91
the time for the first part of the upward-thrown ballโs path. That time
is found using the first kinematic equation with v0 = +v0 and v = โv0:
v = v0 โ gฮt
ฮt =
2.65
v โ v0 โv0 โ v0 2v0
=
=
โg
โg
g
From the time-independent kinematic equation with v0 > 0, ฮy = 2.00 m,
and v = ยฑ1.50 m/s:
v 2 = v02 โ 2g ( ฮy )
v0 = v 2 + 2g ( ฮy ) =
2.66
( ยฑ1.50 m/s) + 2g ( 2.00 m ) = 6.44 m/s
2
(a) To find the distance ฮx traveled by the blood during the acceleration,
apply the time-independent kinematic equation with v0 = 0, v = 1.05
m/s, and a = 22.5 m/s2:
v 2 = v02 + 2aฮx
v 2 โ v02 (1.05 m/s) โ 0
ฮx =
=
= 2.45 ร 10โ2 m = 2.45 cm
2a
2 22.5 m/s2
2
(
)
(b) Solve for the time t required for the blood to reach its peak speed
using the first kinematic equation:
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publicly accessible website, in whole or in part.
Topic 2
92
v = v0 + at
t=
2.67
v โ v0 (1.05 m/s) โ 0
=
= 4.67 ร 10โ2 s
a
22.5 m/s2
When released from rest (v0 = 0), the bill falls freely with a downward
acceleration due to gravity (a = โg = โ9.80 m/s2). Thus, the magnitude of
its downward displacement during Davidโs 0.2 s reaction time will be
1
1
ฮy = v0t + at 2 = 0 + (โ9.80 m/s 2 )(0.2 s)2 = 0.2 m = 20 cm
2
2
This is over twice the distance from the center of the bill to its top edge
(โ8 cm), so David will be unsuccessful.
2.68
(a) The velocity with which the first stone hits the water is
2
mโ
mโ
m
โ
โ
v1 = โ v + 2a(ฮy) = โ โ โ2.00 โ + 2 โ โ9.80 2 โ (โ5.00 m) = โ31.4
โ
โ
sโ
s โ
s
2
01
The time for this stone to hit the water is
t1 =
v1 โ v01 [โ31.4 m/s โ (โ2.00 m/s)]
=
= 3.00 s
a
โ9.80 m/s 2
(b) Since they hit simultaneously, the second stone, which is released
1.00 s later, will hit the water after an flight time of 2.00 s. Thus,
v02 =
ฮy โ at 22 /2 โ50.0 m โ (โ9.80 m/s 2 )(2.00 s)2 /2
=
= โ15.2 m/s
t2
2.00 s
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publicly accessible website, in whole or in part.
Topic 2
93
(c) From part (a), the final velocity of the first stone is v1 = โ31.4 m/s.
The final velocity of the second stone is
v2 = v02 + at2 = โ15.2 m/s + (โ9.80 m/s2)(2.00 s) = โ34.8 m/s
2.69
When the hare wakes up, the tortoise is a distance L > 0 from the finish
line whereas the hare is a distance L + d from the finish line. The hare,
running with constant speed v1, reaches the finish line in a time given by
thare =
L+d
v1
The tortoise, crawling with constant speed v2 < v1, reaches the finish line in
a time given by
ttortoise =
L
v2
The tortoise wins the race if ttortoise < thare, so it follows that
L L+d
<
v2
v1
Rearrange this expression to find a condition on the length L:
Lv1 < Lv2 + v2 d โ L <
v2 d
v1 โ v2
The tortoise wins the race if the length to the finish line, L, satisfies that
inequality.
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publicly accessible website, in whole or in part.
Topic 2
2.70
94
1 2
(a) From ฮy = v0t + at with v0 = 0, we have
2
t=
2(ฮy)
=
a
2(โ23 m)
= 2.2 s
โ9.80 m/s 2
(b) From the time-independent velocity equation, the final speed is:
v 2 = v02 โ 2gฮy
(
)
v = 2gฮy = 2 โ9.80 m/s2 ( โ23 m )
= 21 m/s
Because the man is falling, his final velocity is v = โ21 m/s.
(c) The time it takes for the sound of the impact to reach the spectator is
tsound =
ฮy
23 m
=
= 6.8 ร 10 โ2 s
vsound 340 m/s
so the total elapsed time is ttotal = 2.2 s + 6.8 ร 10โ2 s โ 2.3 s.
2.71
The time required for the stuntman to fall 3.00 m, starting from rest, is
1 2
found from ฮy = v0t + at as
2
1
โ3.00 m = 0 + (โ9.80 m/s 2 )t 2
2
so
t=
2(3.00 m)
= 0.782 s
9.80 m/s 2
(a) With the horse moving with constant velocity of 10.0 m/s, the
horizontal distance is
ฮx = vhorset = (10.0 m/s)(0.782 s) = 7.82 m
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publicly accessible website, in whole or in part.
Topic 2
95
(b) The required time is t = 0.782 s as calculated above.
ยฉ 2018 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a
publicly accessible website, in whole or in part.

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