1.

(a) As the man continues to remain at the same place with respect to the gym, it is
obvious that his net displacement is zero.

(b) In 25 min, the average velocity is

( x2 - x1 ) 0.0 - 0.0
vavg = = = 0.0.
(t2 - t1 ) 25 - 0.0

23
24 CHAPTER 2

2. (a) Using the fact that time = distance/velocity while the velocity is constant, we
find

73.2 m + 73.2 m
vavg = = 1.71 m/s.
1.22 m/s + 2.85 m
73.2 m 73.2 m

(b) Using the fact that distance = vt while the velocity v is constant, we find

. m / s)(60 s) + ( 3.05 m / s)(60 s)

(122
vavg = = 2.14 m / s.
120 s

The graphs are shown below (with meters and seconds understood). The first consists
of two (solid) line segments, the first having a slope of 1.22 and the second having a
slope of 3.05. The slope of the dashed line represents the average velocity (in both
graphs). The second graph also consists of two (solid) line segments, having the same
slopes as before the main difference (compared to the first graph) being that the
stage involving higher-speed motion lasts much longer.
 25

3. Rachel s displacement in reaching the gymnasium is

x x0 =2.80 km 0.00 km = 2.80 km.

and her initial speed is v0 = 6.00 km/h. The time taken for her to reach the gymnasium
is

x x0 2.80 km
t1 = = = 0.467 h = 28.0 min
v0 6.00 km/h

and the time taken for her to reach back home from gymnasium is

2.80 km
t2 = = 0.364 h = 21.82 min.
7.70 km/h

Thus, in 28.0 + 21.82 = 49.82 min, she returns back to home and stops.

During the 7-minute time interval (between 28.0 35.0 min), the return distance
traveled by Rachel is
Ê 7.00 min ˆ
(7.70 km/h) Á ˜ = 0.90 km
Ë 60 min ¯

(a) The magnitude of her average velocity during the time interval 0.00 35.0 min is

displacement 2.80 km - 0.89 km

= = 3.26 km/h
time (35.0 / 60.0) h

(b) During the time interval 0.00 35.0 min, the distance traveled by Rachel is

Ê 7.00 min ˆ
2.80 km + (7.70 km/h) Á ˜ = 2.80 km + 0.90 km = 3.70 km
Ë 60 min ¯

distance 3.70 km
Her average speed is = = 6.34 km/h.
time (35.0 / 60.0) h
26 CHAPTER 2

4. Average speed, as opposed to average velocity, relates to the total distance, as

opposed to the net displacement. The distance D up the hill is, of course, the same as
the distance down the hill, and since the speed is constant (during each stage of the
motion) we have speed = D/t. Thus, the average speed is

Dup + Ddown 2D
=
t up + t down D D
+
vup vdown

which, after canceling D and plugging in vup = 35 km/h and vdown = 60 km/h, yields 44
km/h for the average speed.
 27

5. THINK In this one-dimensional kinematics problem, we re given the position

function x(t), and asked to calculate the position and velocity of the object at a later
time.

EXPRESS The position function is given as x(t) = (3 m/s)t (4 m/s2)t2 + (1 m/s3)t3.

The position of the object at some instant t0 is simply given by x(t0). For the time
interval t1 £ t £ t2 , the displacement is Dx = x(t2 ) - x(t1 ) . Similarly, using Eq. 2-2,
the average velocity for this time interval is

Dx x (t2 ) - x (t1 )
vavg = = .
Dt t2 - t1

ANALYZE

(a) Plugging in t = 1 s into x(t) yields

x(1 s) = (3 m/s)(1 s) (4 m/s2)(1 s)2 + (1 m/s3)(1 s)3 = 0.

(b) With t = 2 s we get x(2 s) = (3 m/s)(2 s) (4 m/s2) (2 s)2 + (1 m/s3)(2 s)3 = 2 m.

(c) With t = 3 s we have x (3 s) = (3 m/s)(3 s) (4 m/s2) (3 s)2 + (1 m/s3)(3 s)3 = 0 m.

(d) Similarly, plugging in t = 4 s gives

x(4 s) = (3 m/s)(4 s) (4 m/s2)(4 s)2 + (1 m/s3) (4 s)3 = 12 m.

(e) The position at t = 0 is x = 0. Thus, the displacement between t = 0 and t = 4 s is

Dx = x(4 s) - x(0) = 12 m - 0 = 12 m.

(f) The position at t = 2 s is subtracted from the position at t = 4 s to give the

displacement: Dx = x(4 s) - x(2 s) = 12 m - ( -2 m) = 14 m . Thus, the average velocity
is
Dx 14 m
vavg = = = 7 m/s.
Dt 2s

(g) The position of the object for the interval 0 £ t £ 4 is plotted below. The straight
line drawn from the point at (t, x) = (2 s, 2 m) to (4 s, 12 m) would represent the
average velocity, answer for part (f).
28 CHAPTER 2

LEARN Our graphical representation illustrates once again that the average velocity
for a time interval depends only on the net displacement between the starting and
ending points.
 29

6. Huber s speed is

v0 = (200 m)/(6.509 s) =30.72 m/s = 110.6 km/h,

where we have used the conversion factor 1 m/s = 3.6 km/h. Since Whittingham beat
Huber by 19.0 km/h, his speed is v1 = (110.6 km/h + 19.0 km/h) = 129.6 km/h, or 36
m/s (1 km/h = 0.2778 m/s). Thus, using Eq. 2-2, the time through a distance of 200 m
for Whittingham is
D x 200 m
Dt = = = 5.554 s.
v1 36 m/s
30 CHAPTER 2

7. The velocity of approach of both cars (car A and car B as shown in the following
figure) that move toward each other is

25 ( 16) = 41 km/h.

The time taken to cover 40 km separation between cars is

Ê 40 ˆ
t = Á ˜h
Ë 41 ¯

(a) The total distance covered by the pigeon is

Ê 40 ˆ
36 ¥ Á ˜ ª 35 km.
Ë 41 ¯

(b) In time t, the first car (car A) moves by a distance

Ê 40 ˆ
Á ˜ ¥16 = 15.61 ª 16 km.
Ë 41 ¯

Therefore, the displacement of the bird with respect to the first car (car A) is about
16 km.
 31

8. The amount of time it takes for each person to move a distance L with speed vs is
Dt = L / vs . With each additional person, the depth increases by one body depth d

(a) The rate of increase of the layer of people is

d d dv (0.25 m)(3.50 m/s)

R= = = s = = 0.50 m/s
D t L / vs L 1.75 m
(b) The amount of time required to reach a depth of D = 5.0 m is
D 5.0 m
t= = = 10 s
R 0.50 m/s
32 CHAPTER 2

9. Converting to seconds, the running times are t1 = 147.95 s and t2 = 148.15 s,

respectively. If the runners were equally fast, then

L1 L2
savg1 = savg 2 fi = .
t1 t 2
From this we obtain
Êt ˆ Ê 148.15 ˆ
L2 - L1 = Á 2 - 1 ˜ L1 = Á - 1 ˜ L1 = 0.00135 L1 ª 1.4 m
Ë t1 ¯ Ë 147.95 ¯

where we set L1 ª 1000 m in the last step. Thus, if L1 and L2 are no different than
about 1.4 m, then runner 1 is indeed faster than runner 2. However, if L1 is shorter
than L2 by more than 1.4 m, then runner 2 would actually be faster.
 33

10. Let vw be the speed of the wind and vc be the speed of the car.

(a) Suppose during time interval t1 , the car moves in the same direction as the wind.
Then the effective speed of the car is given by veff ,1 = vc + vw , and the distance
traveled is d = veff ,1t1 = (vc + vw )t1 . On the other hand, for the return trip during
time interval t2, the car moves in the opposite direction of the wind and the
effective speed would be veff ,2 = vc - vw . The distance traveled is
d = veff ,2t2 = (vc - vw )t2 . The two expressions can be rewritten as
d d
vc + vw = and vc - vw =
t1 t2
1Êd d ˆ
Adding the two equations and dividing by two, we obtain vc = Á + ˜ . Thus,
2 Ë t1 t2 ¯
method 1 gives the car s speed vc a in windless situation.

(b) If method 2 is used, the result would be

d 2d 2d vc2 - vw2 È Ê v ˆ2 ˘
vc¢ = = = = = vc Í1 - Á w ˜ ˙ .
(t1 + t2 ) / 2 t1 + t2 d d vc ÍÎ Ë vc ¯ ˙˚
+
vc + vw vc - vw
The fractional difference is
2
vc - vc¢ Ê vw ˆ
= Á ˜ = (0.0240)2 = 5.76 ¥10-4 .
vc Ë vc ¯
34 CHAPTER 2

11. In 150 s, the pickup vehicle moves a distance of 150.0 × 15.00 = 2250 m. The
total distance covered by the scooter is
1500 m + 2250 m = 3750 m.

The speed at which the scooterist should chase the pickup vehicle is
3750
v0 = = 25.00 m/s.
150.0
 35

12. (a) Let the fast and the slow cars be separated by a distance d at t = 0. If during
the time interval t = L / vs = (12.0 m) /(5.0 m/s) = 2.40 s in which the slow car has
moved a distance of L = 12.0 m , the fast car moves a distance of vt = d + L to join
the line of slow cars, then the shock wave would remain stationary. The condition
implies a separation of

d = vt - L = (25 m/s)(2.4 s) - 12.0 m = 48.0 m.

(b) Let the initial separation at t = 0 be d = 96.0 m. At a later time t, the slow and
the fast cars have traveled x = vs t and the fast car joins the line by moving a
distance d + x . From

x d+x
t= = ,
vs v
we get

vs 5.00 m/s
x= d= (96.0 m) = 24.0 m,
v - vs 25.0 m/s - 5.00 m/s

which in turn gives t = (24.0 m) /(5.00 m/s) = 4.80 s. Since the rear of the
slow-car pack has moved a distance of

Dx = x - L = 24.0 m - 12.0 m = 12.0 m downstream,

the speed of the rear of the slow-car pack, or equivalently, the speed of the shock
wave, is
Dx 12.0 m
vshock = = = 2.50 m/s.
t 4.80 s

(c) Since x > L , the direction of the shock wave is downstream.

36 CHAPTER 2

13. (a) Denoting the travel time and distance from San Antonio to Houston as T and D,
respectively, the average speed is

D (55 km/h)(T/2) + (90 km/h)(T / 2)

savg1 = = = 72.5 km/h
T T

which should be rounded to 73 km/h.

(b) Using the fact that time = distance/speed while the speed is constant, we find

D D
savg2 = = D/2
= 68.3 km/h
T 55 km/h + 90Dkm/h
/2

which should be rounded to 68 km/h.

(c) The total distance traveled (2D) must not be confused with the net displacement
(zero). We obtain for the two-way trip
2D
savg = = 70 km/h.
72.5 km/h + 68.3 km/h
D D

(d) Since the net displacement vanishes, the average velocity for the trip in its
entirety is zero.

(e) In asking for a sketch, the problem is allowing the student to arbitrarily set the
distance D (the intent is not to make the student go to an atlas to look it up); the
student can just as easily arbitrarily set T instead of D, as will be clear in the following
discussion. We briefly describe the graph (with kilometers-per-hour understood for
the slopes): two contiguous line segments, the first having a slope of 55 and
connecting the origin to (t1, x1) = (T/2, 55T/2) and the second having a slope of 90 and
connecting (t1, x1) to (T, D) where D = (55 + 90)T/2. The average velocity, from the
graphical point of view, is the slope of a line drawn from the origin to (T, D). The
graph (not drawn to scale) is depicted below:
 37

14. Using the general property d

dx exp(bx ) = b exp(bx ) , we write

If a concern develops about the appearance of an argument of the exponential ( t)

apparently having units, then an explicit factor of 1/T where T = 1 second can be
inserted and carried through the computation (which does not change our answer).
The result of this differentiation is
v = 16(1 - t ) e - t

with t and v in SI units (s and m/s, respectively). We see that this function is zero
when t = 1 s. Now that we know when it stops, we find out where it stops by
plugging our result t = 1 into the given function x = 16te–t with x in meters. Therefore,
we find x = 5.9 m.
38 CHAPTER 2

15. We have x = 18t + 5.0t 2 .

(a) Instantaneous velocity is calculated as

Dx dx d
v = lim = = (18t + 5.0t 2 )
Dt Æ0 Dt dt dt

That is,
v = 18 + 10.0t (1)

At t = 2.0 s, the instantaneous velocity is

v = 18 + (10 ¥ 2.0) = 18 + 20 = 38 m/s.

(b) Let x1 be the distance of the particle at t = 2.0 s, which is calculated as

x1 = 18(2.0) + 5.0(2.0) 2 = 36 + 20 = 56 m.

Let x2 be the distance of the particle at t = 3.0 s, which is calculated as

x2 = 18(3.0) + 5.0(3.0) 2 = 54 + 45 = 99 m.

Therefore, the average velocity of the particle between t = 2 s and t = 3 s is

calculated as
Dx ( x2 - x1 ) 99 - 56 43
vavg = = = = = 43 m/s.
Dt (t2 - t1 ) 3.0 - 2.0 1.0
 39

16. We use the functional notation x(t), v(t), and a(t) in this solution, where the latter
two quantities are obtained by differentiation:

with SI units understood.

(a) From v(t) = 0 we find it is (momentarily) at rest at t = 0.

(b) We obtain x(0) = 4.0 m.

(c) and (d) Requiring x(t) = 0 in the expression x(t) = 4.0 6.0t2 leads to t = ±0.82 s
for the times when the particle can be found passing through the origin.

(e) We show both the asked-for graph (on the left) as well as the shifted graph that
is relevant to part (f). In both cases, the time axis is given by 3 £ t £ 3 (SI units
understood).

(f) We arrived at the graph on the right (shown above) by adding 20t to the x(t)
expression.

(g) Examining where the slopes of the graphs become zero, it is clear that the shift
causes the v = 0 point to correspond to a larger value of x (the top of the second
curve shown in part (e) is higher than that of the first).
40 CHAPTER 2

17. We use Eq. 2-2 for average velocity and Eq. 2-4 for instantaneous velocity, and
work with distances in centimeters and times in seconds.

(a) We plug into the given equation for x for t = 2.00 s and t = 3.00 s and obtain x2 =
21.75 cm and x3 = 50.25 cm, respectively. The average velocity during the time
interval 2.00 £ t £ 3.00 s is
Dx 50.25 cm - 2175
. cm
vavg = =
Dt 3.00 s - 2.00 s

which yields vavg = 28.5 cm/s.

(b) The instantaneous velocity is v = dx

dt = 4.5t 2 , which, at time t = 2.00 s, yields v =
(4.5)(2.00)2 = 18.0 cm/s.

(c) At t = 3.00 s, the instantaneous velocity is v = (4.5)(3.00)2 = 40.5 cm/s.

(d) At t = 2.50 s, the instantaneous velocity is v = (4.5)(2.50)2 = 28.1 cm/s.

(e) Let tm stand for the moment when the particle is midway between x2 and x3 (that is,
when the particle is at xm = (x2 + x3)/2 = 36 cm). Therefore,

xm = 9.75 + 15
. t m3 fi tm = 2.596

in seconds. Thus, the instantaneous speed at this time is v = 4.5(2.596)2 = 30.3 cm/s.

(f) The answer to part (a) is given by the slope of the straight line between t = 2 and t
= 3 in this x-vs-t plot. The answers to parts (b), (c), (d), and (e) correspond to the
slopes of tangent lines (not shown but easily imagined) to the curve at the appropriate
points.
 41

18. (a) Taking derivatives of x(t) = 12t2 2t3 we obtain the velocity and the
acceleration functions:
v(t) = 24t 6t2 and a(t) = 24 12t

with length in meters and time in seconds. Plugging in the value t = 3.5 s yields
x(3.5) = 61 m .

(b) Similarly, plugging in the value t = 3.5 s yields v(3.5) = 11 m/s.

(c) For t = 3.5 s, a(3.5) = 18 m/s2.

(d) At the maximum x, we must have v = 0; eliminating the t = 0 root, the velocity
equation reveals t = 24/6 = 4 s for the time of maximum x. Plugging t = 4 into
the equation for x leads to x = 64 m for the largest x value reached by the particle.

(e) From (d), we see that the x reaches its maximum at t = 4.0 s.

(f) A maximum v requires a = 0, which occurs when t = 24/12 = 2.0 s. This, inserted
into the velocity equation, gives vmax = 24 m/s.

(g) From (f), we see that the maximum of v occurs at t = 24/12 = 2.0 s.

(h) In part (e), the particle was (momentarily) motionless at t = 4 s. The acceleration at
that time is readily found to be 24 12(4) = 24 m/s2.

(i) The average velocity is defined by Eq. 2-2, so we see that the values of x at t = 0
and t = 3 s are needed; these are, respectively, x = 0 and x = 54 m (found in part
(a)). Thus,
54 - 0
vavg = = 18 m/s.
3- 0
42 CHAPTER 2

19. THINK In this one-dimensional kinematics problem, we re given the speed of a

particle at two instants and asked to calculate its average acceleration.

EXPRESS We represent the initial direction of motion as the +x direction. The

average acceleration over a time interval t1 £ t £ t2 is given by Eq. 2-7:

Dv v(t2 ) - v(t1 )
aavg = = .
Dt t2 - t1

ANALYZE Let v1 = +18 m/s at t1 = 0 and v2 = 30 m/s at t2 = 2.4 s. Using Eq. 2-7
we find
v(t ) - v(t1 ) (-30 m/s) - (+1 m/s)
aavg = 2 = = - 20 m/s 2 .
t2 - t1 2.4 s - 0

LEARN The average acceleration has magnitude 20 m/s2 and is in the opposite
direction to the particle s initial velocity. This makes sense because the velocity of the
particle is decreasing over the time interval. With t1 = 0 , the velocity of the particle
as a function of time can be written as

v = v0 + at = (18 m/s) - (20 m/s 2 )t .

 43

20. The position of a particle is x = 25t - 6t 3 .

(a) The particle s velocity is
dx
v= = (25 - 18t 2 ) m/s.
dt
If v = 0 m/s, we have

18t 2 = 25.

Therefore,

25 Ê 5 ˆ
t2 = fi t = ±Á ˜ s = ±1.2 s.
18 Ë3 2 ¯

That is, the particle s velocity is zero when t = ±1.2 s.

(b) The instantaneous acceleration of the particle is

dv d 2
a= = ( x) = (0 - 36t ) m/s 2 .
dt dt 2
That is,

a = 0 m/s 2 or - 36t = 0 fi t = 0 s.

Therefore, at t = 0 s, the acceleration of the particle is zero.

(c) It is obvious that a(t) = 36t is negative for t > 0.
(d) The acceleration a(t) = 36t is positive for t < 0.
(e) The graphs are shown in Figs. (a), (b) and (c) as follows (SI units are understood):
44 CHAPTER 2

21. The initial velocity of the car is

130 ¥ 5.00 650
v0 = 130 km/h = m/s = m/s = 36.1 m/s.
18.0 18.0
The displacement of the car is
x x0 = 210 m 0.00 m = 210 m.

(a) Using one of the equations of motion,

v 2 = v02 + 2 a ( x - x0 ) ,

we get the acceleration, a, as

v2 - v02
a= .
2( x x0 )

Substituting the values in this equation, we get the uniform retardation, that is,
uniform deceleration of the car as follows:

v2 - v02 0 2 - (36.1) 2 1303.21

a= = =- = -3.10 m/s 2 .
2( x x0 ) 2(210) 420

Thus, the magnitude is 3.10 m/s2.

(b) Using one of the equations of motion, v = v0 + at , we calculate the time taken
by the car to stop as follows:
v - v0 0 - 36.1
t= = = 11.64 s ª 11.6s.
a -3.1
 45

22. In this solution, we make use of the notation x(t) for the value of x at a particular t.
The notations v(t) and a(t) have similar meanings.

(a) Since the unit of ct2 is that of length, the unit of c must be that of length/time2, or
m/s2 in the SI system.

(b) Since bt3 has a unit of length, b must have a unit of length/time3, or m/s3.

(c) When the particle reaches its maximum (or its minimum) coordinate its velocity is
zero. Since the velocity is given by v = dx/dt = 2ct 3bt2, v = 0 occurs for t = 0 and
for
2c 2(4.0 m/s 2 )
t= = = 1.3 s .
3b 3(2.0 m/s3 )

For t = 0, x = x0 = 0 and for t = 1.3 s, x = 2.3 m > x0. Since we seek the maximum, we
reject the first root (t = 0) and accept the second (t = 1s).

(d) In the first 4 s the particle moves from the origin to x = 1.0 m, turns around, and
goes back to

x(4 s) = (4.0 m/s 2 )(4.0 s) 2 - (2.0 m/s3 )(4.0 s)3 = - 64 m .

The total path length it travels is 1.3 m + 1.3 m + 64 m = 67 m.

(e) Its displacement is Dx = x2 x1, where x1 = 0 and x2 = 64 m. Thus, Dx = -64 m .

The velocity is given by v = 2ct 3bt2 = (8.0 m/s2)t (6.0 m/s3)t2.

(f) Plugging in t = 1 s, we obtain

v (1.0) = 2.0 m/s

(g) Similarly, v (2.0) = -8.0 m/s

(h) v (3.0) = -30 m/s

(i) v (4.0) = -64 m/s

The acceleration is given by a = dv/dt = 2c 6b = 8.0 m/s2 (12.0 m/s3)t.

(j) Plugging in t = 1 s, we obtain

a (1.0) = -4.0 m/s2

(k) a (2.0) = -16 m

(l) a (3.0) = -28 m

(m) a (4.0) = -40 m

46 CHAPTER 2

23. It is given that v0 = 0. Let the constant acceleration be a. Distance covered

during the fifth second is the distance moved between t = 4.00 s and t = 5.00 s. The
velocity of the object at t = 4.00 s is used as the initial velocity for its further motion
at 4.00 s is
v4 = v0 + at fi v4 = (4.00) a.

Therefore, the distance covered between t = 4.00 s and t = 5.00 s is

1 1
( x - x0 )45 = v4t + at 2 = (4.00)a + a = (4.50)a.
2 2
Given that t = 1.00 s between t = 4.00 s and t = 5.00 s. The distance covered up to
5.00 s is
1 È1 ˘ 1
( x - x0 )5 = v0 t + at 2 = 0 + Í (a)(5.00)2 ˙ = a (25.0) = (12.5)a.
2 Î2 ˚ 2
Given that t = 5.00 s, the required ratio is
( x - x0 ) 45 (4.50) a
= = 0.36.
( x - x0 ) 5 (12.5) a
 47

24. In this problem we are given the initial and final speeds, and the displacement, and
are asked to find the acceleration. We use the constant-acceleration equation given in
Eq. 2-16, v2 = v20 + 2a(x x0).

(a) Given that v0 = 0 , v = 1.6 m/s, and Dx = 5.0µ m, the acceleration of the spores
during the launch is
v 2 - v02 (1.6 m/s) 2
a= = = 2.56 ¥105 m/s 2 = 2.6 ¥104 g
2x 2(5.0 ¥10 -6 m)

(b) During the speed-reduction stage, the acceleration is

v 2 - v02 0 - (1.6 m/s) 2

a= = = -1.28 ¥103 m/s 2 = -1.3 ¥102 g
2x 2(1.0 ¥10-3 m)

The negative sign means that the spores are decelerating.

48 CHAPTER 2

25. From the equation of motion

v = v0 + at ,
we get

2.8 = v0 + 2.5a (1)

Using another equation of motion, v 2 - v02 = 2 a ( x - x0 ) , we get

(2.8)2 - v02 = 2a(8.0 - 2.0) fi (2.8 + v0 ) (2.8 - v0 ) = 12a (2)

Dividing Eq. (2) by Eq. (1) gives

(2.8 + v0 )(2.8 - v0 ) 12a
=
(2.8 - v0 ) 2.5a
which leads to
12 4.8
2.8 + v0 = = 4.8 fi v0 = = 2.0 m/s.
2.5 2.8
Substituting the value of v0 in Eq. (1), we get the acceleration during the given time
interval as follows:
2.8 = v0 + 2.5a
fi 2.8 - 2 = 2.5a
0.8 8 ¥ 10
fia= = = 0.32 m/s 2 .
2.5 10 ¥ 25
 49

26. The constant-acceleration condition permits the use of Table 2-1.

(a) Setting v = 0 and x0 = 0 in v 2 = v02 + 2a ( x - x0 ) , we find

1 v02 1 (6.00 ¥ 106 )2

x=- =- = 0.144 m .
2 a 2 -1.25 ¥ 1014

Since the muon is slowing, the initial velocity and the acceleration must have
opposite signs.

(b) Below are the time plots of the position x and velocity v of the muon from the
moment it enters the field to the time it stops. The computation in part (a) made no
reference to t, so that other equations from Table 2-1 (such as v = v0 + at and
x = v0 t + 12 at 2 ) are used in making these plots.
50 CHAPTER 2

27. We have v0 = 0; ( x - x0 ) = 2.0 cm; t = (5.0 ¥ 10 -3 ) s. Substituting these values in the

equation of motion
1
( x - x0 ) = v0 t + at 2 ,
2
we get
Ê 2.00 ˆ È 1 ˘
Á ˜m = 0 + Í ¥ a (5.00 ¥ 10-3 ) 2 ˙
Ë 100 ¯ Î 2.00 ˚

The magnitude of the given acceleration is obtained as follows:

Ê 4.00 ˆ Ê 4.00 ˆ
a=Á -6 ˜
m/s 2 = Á ¥ 10 4 ˜ m/s2 = (1.6 ¥ 103 ) m/s2 .
Ë 100 ¥ 25.0 ¥ 10 ¯ Ë 25.0 ¯
 51

28. We take +x in the direction of motion, so v0 = +27.2 m/s and a = 4.92 m/s2. We
also take x0 = 0.

(a) The time to come to a halt is found using Eq. 2-11:

27.2 m/s
0 = v0 + at fi t = = 5.53 s
4.92 m/s 2

(b) Although several of the equations in Table 2-1 will yield the result, we choose Eq.
2-16 (since it does not depend on our answer to part (a)).
( 27.2 m/s )
2

0 = v + 2 ax fi x = -
2
0 = 75.2 m
2( -4.92 m/s 2 )

(c) Using these results, we plot v0t + 12 at 2 (the x graph, shown next, on the left) and
v0 + at (the v graph, on the right) over 0 £ t £ 5 s, with SI units understood.
52 CHAPTER 2

29. We assume the periods of acceleration (duration t1) and deceleration (duration t2)
are periods of constant a so that Table 2-1 can be used. Taking the direction of motion
to be +x then a1 = +1.22 m/s2 and a2 = 1.22 m/s2. We use SI units so the velocity at t
= t1 is v = 305/60 = 5.08 m/s.

(a) We denote Dx as the distance moved during t1, and use Eq. 2-16:

(5.08 m/s)2
v 2 = v02 + 2 a1Dx fi Dx = = 10.59 m ª 10.6 m.
2(1.22 m/s 2 )
(b) Using Eq. 2-11, we have
v - v0 5.08 m/s
t1 = = = 4.17 s.
a1 1.22 m/s 2

The deceleration time t2 turns out to be the same so that t1 + t2 = 8.33 s. The distances
traveled during t1 and t2 are the same so that they total to 2(10.59 m) = 21.18 m. This
implies that for a distance of 190 m 21.18 m = 168.82 m, the elevator is traveling at
constant velocity. This time of constant velocity motion is

168.82 m
t3 = = 33.21 s.
5.08 m / s

Therefore, the total time is 8.33 s + 33.21 s ª 41.5 s.

 53

30. We choose the positive direction to be that of the initial velocity of the car
(implying that a < 0 since it is slowing down). We assume the acceleration is constant
and use Table 2-1.

(a) Substituting v0 = 146 km/h = 40.6 m/s, v = 90 km/h = 25 m/s, and a = 5.2 m/s2
into v = v0 + at, we obtain

25 m/s - 40.6 m/s

t= = 3.0 s .
-5.2 m/s 2

(b) We take the car to be at x = 0 when the brakes

are applied (at time t = 0). Thus, the coordinate
of the car as a function of time is given by

1
• x = ( 30.6 m/s ) t +
2
( -5.2 m/s2 ) t 2
in SI units. This function is plotted from t = 0 to
t = 2.5 s on the graph to the right. We have not
shown the v-vs-t graph here; it is a descending
straight line from v0 to v.
54 CHAPTER 2

31. We have the initial velocity of the rocket, v0 = 0; acceleration of the rocket, a =
10.0 m/s2. Referring to the following figure, which depicts the situation, the velocity
at A is calculated as follows: From the equation of motion,

v 2 = v02 + 2 a ( x - x0 ) ,

we get
v 2 - v02 = 2a( x - x0 )

fi v2 - 02 = 2(10.0 m/s2 ¥ 500 m)

fi v2 = 10, 000 fi v = 100 m/s.

After the engine of the rocket cuts off, the rocket rises against gravity with initial
velocity of 100 m/s. Therefore,
u 2 100 ¥ 100
h1 = = = 510.2 m.
2g 2 ¥ 9.8

Therefore, the maximum altitude the rocket reaches is

H = h + h1 = 500 m + 510.2 m = 1010.2 m = 1.01 km.

 55

32. The acceleration is found from Eq. 2-11 (or, suitably interpreted, Eq. 2-7).

In terms of the gravitational acceleration g, this is expressed as a multiple of 9.8 m/s2

as follows:
Ê 202.4 m/s 2 ˆ
a=Á 2 ˜
g = 21g .
Ë 9.8 m/s ¯
56 CHAPTER 2

33. The position of the stone is given by

1
y = 60 m - (20 m/s)t - (9.8 m/s 2 )t 2
2
Solving the quadratic equation with y = 0, we obtain t = +2.01 s or - 6.09 s. Since a
negative value implies that the stone reaches the ground before it is released, we
consider the positive value, that is, +2.0 s.
 57

34. Let d be the 220 m distance between the cars at t = 0, and v1 be the 20 km/h = 50/9
m/s speed (corresponding to a passing point of x1 = 44.5 m) and v2 be the 40 km/h
=100/9 m/s speed (corresponding to a passing point of x2 = 77.9 m) of the red car.
We have two equations (based on Eq. 2-17):
1
d x1 = vo t1 + 2
a t12 where t1 = x1 ⁄ v1

1
d x2 = vo t2 + 2
a t22 where t2 = x2 ⁄ v2

We simultaneously solve these equations and obtain the following results:

(a) The initial velocity of the green car is vo = -8.7 m/s. The negative indicates that
the green car initially moves toward the green car.

(b) The corresponding acceleration of the car is a = -3.3 m/s2.

58 CHAPTER 2

35. Using v 2 = v02 + 2a( x - x0 ) , we find the acceleration of the particle to be

v 2 - v02 (6.00 ¥ 107 m/s) 2 - (4.00 ¥ 105 m/s) 2

a= = = 6.00 ¥ 1016 m/s 2 .
2( x - x0 ) 2(0.0300 m)

The time interval at which the electron accelerates is obtained as follows: From the
equation of motion, v = v0 + at, we have the required time interval as

v - v0 6.00 ¥ 107 m/s - 4.00 ¥ 105 m/s

t= = = 9.93 ¥ 10-10 s = 0.993 ns.
a 6.00 ¥ 1016 m/s2
 59

36. (a) Equation 2-15 is used for part 1 of the trip:

1
Dx1 = vo1 t1 + 2
a1 t12

900
where vo1 = 0, a 1 = 2.75 m/s2 and Dx1 = 4
m = 225 m. Solving for t1, we obtain

2( Dx1 ) 2(225 m)
t1 = = = 12.79 s
a1 2.75 m/s 2

The speed attained during part 1 of the trip is

v1 f = a1t1 = 2a1 (Dx1 ) = 2(2.75 m/s 2 )(225 m) = 35.2 m/s.

With v20 = v1 f , the time it takes for the car to come to rest with a constant acceleration

of a 2 = -0.917 m/s2 is given by v2 f = v20 + a2t2 = 0, or

v2 f - v20 0 - 35.2 m/s

t2 = = = 38.36 s
a2 -0.917 m/s 2

The total travel time is t = t1 + t2 = 12.79 s + 38.36 s = 51.2 s.

(b) The maximum speed is attained at the end of the first part of the trip:
v1 f = 35.2 m/s.
60 CHAPTER 2

37. (a) From the figure, we see that x0 = 2.0 m. From Table 2-1, we can apply

x x0 = v0t + 1
2 at2

with t = 1.0 s, and then again with t = 2.0 s. This yields two equations for the two
unknowns, v0 and a:
1
0.0 - ( -2.0 m ) = v0 (1.0 s ) + a (1.0 s )
2

2
1
6.0 m - ( -2.0 m ) = v0 ( 2.0 s ) + a ( 2.0 s ) .
2

Solving these simultaneous equations yields the results v0 = 0 and a = 4.0 m/s2.

(b) The fact that the answer is positive tells us that the acceleration vector points in
the +x direction.
 61

38. We assume the train accelerates from rest ( v0 = 0 and x0 = 0 ) at

a1 = + 1.34 m / s until the midway point and then decelerates at a2 = - 1.34 m / s2
2

until it comes to a stop at the next station. The velocity at the midpoint is v1,
which occurs at x1 = 806/2 = 403m.

(a) Equation 2-16 leads to

v12 = v02 + 2a1 x1 fi v1 = 2 (1.34 m/s 2 ) ( 440 m ) = 34.3 m/s.

(b) The time t1 for the accelerating stage is (using Eq. 2-15)

1 2 2 ( 440 m )
x1 = v0t1 + a1t1 fi t1 = = 25.6 s .
2 1.34 m/s 2

Since the time interval for the decelerating stage turns out to be the same, we double
this result and obtain t = 51.2 s for the travel time between stations.

(c) With a dead time of 20 s, we have T = t + 20 = 71.2 s for the total time between
start-ups. Thus, Eq. 2-2 gives
880 m
vavg = = 12.4 m/s .
71.2 s

(d) The graphs for x, v and a as a function of t are shown below. The third graph, a(t),
consists of three horizontal steps one at 1.34 m/s2 during 0 < t < 24.53 s, and
2
the next at 1.34 m/s during 24.53 s < t < 49.1 s and the last at zero during the dead
time 49.1 s < t < 69.1 s).
62 CHAPTER 2
 63

39. Let t be the time taken by the traffic cop to overtake the speeding car. The total
time (reaction plus chase) in which the car moves is (t + 1.0) s. Using the equation of
motion
1
( x - x0 ) = v0 t + at 2 ,
2
we get

Ê1 ˆ
46(t + 1) = 0(t ) + Á ¥ 4 ¥ t 2 ˜
Ë2 ¯
fi 46t + 46 = 2t 2
fi t 2 - 46t - 46 = 0,

which is simplified further to get the time that it takes to the cop to overtake the
speeding car as follows:

46 ± 2116 + 368 46 ± 2484 46 ± 49.84 95.84

t= = = = = 23.96 ª 24 s.
4 4 4 4
64 CHAPTER 2

40. We take the direction of motion as +x, so a = 5.18 m/s2, and we use SI units, so
v0 = 55(1000/3600) = 15.28 m/s.

(a) The velocity is constant during the reaction time T, so the distance traveled during
it is
dr = v0T (15.28 m/s) (0.75 s) = 11.46 m.

We use Eq. 2-16 (with v = 0) to find the distance db traveled during braking:

(15.28 m/s) 2
v 2 = v02 + 2 ad b fi d b = -
2 ( -5.18 m/s 2 )

which yields db = 22.53 m. Thus, the total distance is d r + d b = 34.0 m, which

means that the driver is able to stop in time. And if the driver were to continue at
v0, the car would enter the intersection in t = (40 m)/(15.28 m/s) = 2.6 s, which is
(barely) enough time to enter the intersection before the light turns, which many
people would consider an acceptable situation.

(b) In this case, the total distance to stop (found in part (a) to be 34 m) is greater than
the distance to the intersection, so the driver cannot stop without the front end of
the car being a couple of meters into the intersection. And the time to reach it at
constant speed is 32/15.28 = 2.1 s, which is too long (the light turns in 1.8 s). The
driver is caught between a rock and a hard place.
 65

41. (a) Using the equation of motion v = v0 + at , we get the acceleration of the jet
when it stops in 3.0 s as follows:
Ê 64 ˆ
0 = 64 m/s + a(3.0 s) fi a = Á - ˜ m/s2 .
Ë 3 ¯
The magnitude of the acceleration is about 21 m/s2.

(b) Using the equation of motion

1 2
( x - x0 ) = v0 t + at ,
2
we get final position of the jet when it touches the position at xi = 0 as follows:

1 2 1
x = v0t + at = (64 m/s)(3.0 s) + (-21.33 m/s 2 )(3.0 s)2 = 192 m - 96 m = 96 m.
2 2
66 CHAPTER 2

42. (a) Note that 120 km/h is equivalent to 33.3 m/s. During a two-second interval,
you travel 66.67 m. The decelerating police car travels (using Eq. 2-15)

1
Dx p = (33.3 m/s)(2.0 s) - (5.0 m/s 2 )(2.0 s) 2 = 56.67 m
2

which is 10.0 m less than that by your car. The initial distance between cars was 25 m,
this means the gap between cars is now 15.0 m.

(b) First, we add 0.4 s to the considerations of part (a). During the 2.4 s interval, you
travel 80.0 m, while the decelerating police car travels
1
Dx p = (33.3 m/s)(2.4 s) - (5.0 m/s 2 )(2.4 s) 2 = 65.6 m
2
which is 14.4 m less than that by your car.

The initial distance between cars of 25 m has now become 25 m 14.4 m = 10.6 m.
The speed of the police car at the instant you begin to brake is 33.3 m/s (5.00
m/s2)(2.4 s) = 21.33 m/s. Collision occurs at time t when xyou = xpolice (we choose
coordinates such that your position is x = 0 and the police car s position is x = 10.6 m
at t = 0). Eq. 2-15 becomes, for each car:
1
xpolice = 10.6 m + (21.3 m/s)t 2
(5.00 m/s2)t2
1
xyou = (33.3 m/2)t 2
(5.00 m/s2)t2 .

Subtracting equations, we find

10.6 m = (33.3 m/s 21.3 m/s)t fi t = 0.883 s

At that time your speed is (33.3 m/s) (5.00 m/s2)(0.883 s) ª 28.9 m/s, or 104 km/h.
 67

43. In this solution we elect to wait until the last step to convert to SI units. Constant
acceleration is indicated, so use of Table 2-1 is permitted. We start with Eq. 2-17 and
denote the train s initial velocity as vt and the locomotive s velocity as vl (which is
also the final velocity of the train, if the rear-end collision is barely avoided). We note
that the distance Dx consists of the original gap between them, D, as well as the
forward distance traveled during this time by the locomotive vl t . Therefore,
vt + v l D x D + vl t D
= = = + vl .
2 t t t

We now use Eq. 2-11 to eliminate time from the equation. Thus,

which leads to

Hence,

which we convert as follows:

so that its magnitude is |a| = 0.994 m/s2. A graph is

shown here for the case where a collision is just
avoided (x along the vertical axis is in meters and t
along the horizontal axis is in seconds). The top
(straight) line shows the motion of the locomotive
and the bottom curve shows the motion of the
passenger train.

The other case (where the collision is not quite

avoided) would be similar except that the slope of
the bottom curve would be greater than that of the
top line at the point where they meet.
68 CHAPTER 2

44. We neglect air resistance, which justifies setting a = g = 9.8 m/s2 (taking down
as the y direction) for the duration of the motion. We are allowed to use Table 2-1
(with Dy replacing Dx) because this is constant acceleration motion. The ground level
is taken to correspond to the origin of the y axis.

(a) Using y = v0t - 12 gt 2 , with y = 0.558 m and t = 0.200 s, we find

y + gt 2 / 2 0.558 m + (9.8 m/s 2 ) (0.200 s)2 / 2
v0 = = = 3.77 m/s .
t 0.200 s

(b) The velocity at y = 0.558 m is

v = v0 - gt = 3.77 m/s - (9.8 m/s 2 ) (0.200 s) = 1.81 m/s .

(b) Using v 2 = v02 - 2 gy (with different values for y and v than before), we solve for
the value of y corresponding to maximum height (where v = 0).
v 2 (3.77 m/s) 2
y= 0 = = 0.725 m.
2 g 2(9.8 m/s 2 )

Thus, the armadillo goes 0.725 0.558 = 0.167 m higher.

 69

45. Let the journey last for t seconds from the top of the tower. The height of the
tower is given by
1 2
H = gt .
2
The distance covered during the tth second is
1 2 1 Ê 1ˆ
Dy = gt - g (t - 1) 2 = g Á t - ˜
2 2 Ë 2¯
Since Dy = 9 H / 25, we have
Ê 1ˆ 9 9 Ê1 ˆ 9 2
g Át - ˜ = H = Á gt 2 ˜ = gt
Ë 2 ¯ 25 25 Ë 2 ¯ 50
which can be simplified to give
9t 2 - 50t + 25 = (9t - 5)(t - 5) = 0

Thus, we find t = 5.0 s. Note that the other solution t = (5 / 9) s is not possible since
t > 0.5 . We conclude that
1 2 1
H= gt = (9.8 m/s 2 )(5.0 s) 2 = 122.5 m.
2 2
70 CHAPTER 2

46. Neglect of air resistance justifies setting a = g = 9.8 m/s2 (where down is our y
direction) for the duration of the fall. This is constant acceleration motion, and we
may use Table 2-1 (with Dy replacing Dx).

(a) Using Eq. 2-16 and taking the negative root (since the final velocity is downward),
we have
v = - v02 - 2 g Dy = - 0 - 2(9.8 m/s 2 )(-1800 m) = -188 m/s .

Its magnitude is therefore 188 m/s.

(b) No, but it is hard to make a convincing case without more analysis. We estimate
the mass of a raindrop to be about a gram or less, so that its mass and speed (from
part (a)) would be less than that of a typical bullet, which is good news. But the
fact that one is dealing with many raindrops leads us to suspect that this scenario
poses an unhealthy situation. If we factor in air resistance, the final speed is
smaller, of course, and we return to the relatively healthy situation with which we
are familiar.
 71

47. As the balloon is ascending at a velocity of 14 m/s, the initial velocity of the
packet is v0 = 14 m/s. The net displacement traveled by the packet is given by S = -98
m. As the packet reaches the ground, let v be the velocity of the packet. Here, a = -g =
-9. 8 m/s2.

(a) As the packet reaches the ground, the velocity of the packet is calculated as
follows: From the equation of motion

v 2 = v02 + 2a( x - x0 ),

we have

v 2 = (14) 2 + [2 ¥ -9.8 ¥ 98]

(b) That is, v = -46.008 m/s ª - 46 m/s. We choose the negative root because the
velocity is downward.

The time taken by the packet to reach the ground is obtained as follows:
v = v0 + at
-46.008 = 14 - 9.8t
60.08
fit = = 6.1 m/s.
9.8
72 CHAPTER 2

48. We neglect air resistance, which justifies setting a = g = 9.8 m/s2 (taking down
as the y direction) for the duration of the fall. This is constant acceleration motion,
which justifies the use of Table 2-1 (with Dy replacing Dx).

(a) Noting that Dy = y y0 = 30 m, we apply Eq. 2-15 and the quadratic formula
(Appendix E) to compute t:
1 2 v0 ± v02 - 2 gDy
Dy = v0t - gt fi t =
2 g

which (with v0 = 15 m/s since it is downward) leads, upon choosing the positive root
(so that t > 0), to the result:

-15 m/s + (-15 m/s)2 - 2(9.8 m/s 2 )(-30 m)

t= = 1.38 s.
9.8 m/s 2

(b) Enough information is now known that any of the equations in Table 2-1 can be
used to obtain v; however, the one equation that does not use our result from part (a)
is Eq. 2-16:
v = v02 - 2 g Dy = 28.5 m/s

where the positive root has been chosen in order to give speed (which is the
magnitude of the velocity vector).
 73

49. THINK In this problem a package is dropped from a hot-air balloon which is
ascending vertically upward. We analyze the motion of the package under the
influence of gravity.

EXPRESS We neglect air resistance, which justifies setting a = g = 9.8 m/s2

(taking down as the y direction) for the duration of the motion. This allows us to use
Table 2-1 (with Dy replacing Dx):
v = v0 - gt (2 - 11)
1
y - y0 = v0t - gt 2 (2 - 15)
2
v 2 = v02 - 2 g ( y - y0 ) (2 - 16)

We place the coordinate origin on the ground and note that the initial velocity of the
package is the same as the velocity of the balloon, v0 = +12 m/s and that its initial
coordinate is y0 = +80 m. The time it takes for the package to hit the ground can be
found by solving Eq. 2-15 with y = 0.

ANALYZE (a) We solve 0 = y = y0 + v0t - 12 gt 2 for time using the quadratic

formula (choosing the positive root to yield a positive value for t):

v0 + v02 + 2 gy0 12 m/s + (12 m/s) + 2 ( 9.8 m/s ) ( 80 m )

2 2

t= = = 5.4 s .
g 9.8 m/s 2

(b) The speed of the package when it hits the ground can be calculated using Eq. 2-11.
The result is

v = v0 - gt = 12 m/s - (9.8 m/s 2 )(5.447 s) = -41.38 m/s .

Its final speed is about 41 m/s.

LEARN Our answers can be readily verified by using Eq. 2-16 which was not used in
either (a) or (b). The equation leads to

v = - v02 - 2 g ( y - y0 ) = - (12 m/s)2 - 2(9.8 m/s 2 )(0 - 80 m) = -41.38 m/s

which agrees with that calculated in (b).

74 CHAPTER 2

50. (a) The horizontal component of the velocity of the skier remains unchanged,
that is,
vx = v0 = 29.4 m/s.
(b) The vertical component of the velocity of the skier changes with time. Using the
equation of motion
v = v0 + at,
we get

v y = 0 + gt = 0 + ( - 9.8 ¥ 3) = -29.4 m/s.

Hence, we conclude that the magnitude of both horizontal and vertical components
of the skier s velocity are same.
 75

51. For the displacement during the acceleration, we have

1
( x - x0 ) = - ¥ 39.2 = -19.6 m.
2
From the equation of motion

v 2 = v02 + 2 a ( x - x0 ),

we have

v 2 = 0 + (2 ¥ (-9.8) ¥ (-19.6))
fi v = -19.6 m/s,

where v is the velocity of the melon, which makes the melon to hit the ground. We
take the negative root because the melon is moving downward.
76 CHAPTER 2

52. The full extent of the bolt s fall is given by

1
y – y0 = 2
g t2

where y y0 = 100 m (if upward is chosen as the positive y direction). Thus the time
for the full fall is found to be t = 4.52 s. The first 80% of its free-fall distance is given
by 80 = g t2/2, which requires time t = 4.04 s.

(a) Thus, the final 20% of its fall takes t t = 0.48 s.

(b) We can find that speed using v = -gt. Therefore, |v| = 40 m/s, approximately.

(c) Similarly, vfinal = - g t fi |vfinal | = 44 m/s.

 77

53. THINK This problem involves two objects: a key dropped from a bridge, and a
boat moving at a constant speed. We look for conditions such that the key will fall
into the boat.

EXPRESS The speed of the boat is constant, given by vb = d/t, where d is the distance
of the boat from the bridge when the key is dropped (12 m) and t is the time the key
takes in falling.

To calculate t, we take the time to be zero at the instant the key is dropped, we
compute the time t when y = 0 using y = y0 + v0t - 12 gt 2 , with y0 = 45 m. Once t is
known, the speed of the boat can be readily calculated.

ANALYZE Since the initial velocity of the key is zero, the coordinate of the key is

given by y0 = 12 gt 2 . Thus, the time it takes for the key to drop into the boat is

2 y0 2(45 m)
t= = = 3.03 s .
g 9.8 m/s 2
12 m
Therefore, the speed of the boat is vb = = 4.0 m/s.
3.03 s

LEARN From the general expression

d d g
vb = = =d ,
t 2 y0 / g 2 y0

we see that vb : 1/ y0 . This agrees with our intuition that the lower the height from

which the key is dropped, the greater the speed of the boat in order to catch it.
78 CHAPTER 2

54. (a) We neglect air resistance, which justifies setting a = g = 9.8 m/s2 (taking
down as the y direction) for the duration of the motion. We are allowed to use Eq.
2-15 (with Dy replacing Dx) because this is constant acceleration motion. We use
primed variables (except t) with the first stone, which has zero initial velocity, and
unprimed variables with the second stone (with initial downward velocity v0, so that
v0 is being used for the initial speed). SI units are used throughout.
1
Dy¢ = 0 ( t ) - gt 2
2
1
Dy = ( -v0 )( t - 1) - g ( t - 1)
2

2
Since the problem indicates Dy’ = Dy = 53.6 m, we solve the first equation for t:

2(-Dy¢) 2(53.6 m)
t= = = 3.307 s
g 9.8 m/s 2

Using this result to solve the second equation for the initial speed of the second stone:

1
-53.6 m = ( -v0 ) ( 2.307 s ) -
2
( 9.8 m/s 2 ) ( 2.307 s )
2

we obtain v0 = 11.9 m/s.

(b) The velocity of the stones are given by

d ( Dy ¢) d (Dy )
v ¢y = = - gt , vy = = -v0 - g (t - 1)
dt dt

The plot is shown below:

 79

55. THINK The free-falling moist-clay ball strikes the ground with a non-zero speed,
and it undergoes deceleration before coming to rest.

EXPRESS During contact with the ground its average acceleration is given by
Dv
a avg = , where Dv is the change in its velocity during contact with the ground and
Dt
Dt = 20.0 ¥10-3 s is the duration of contact. Thus, we must first find the velocity of the
ball just before it hits the ground (y = 0).

ANALYZE
(a) Now, to find the velocity just before contact, we take t = 0 to be when it is dropped.
Using Eq. 2-16 with y0 = 15.0 m , we obtain

v = - v02 - 2 g ( y - y0 ) = - 0 - 2(9.8 m/s 2 )(0 - 15 m) = -17.15 m/s

where the negative sign is chosen since the ball is traveling downward at the moment
of contact. Consequently, the average acceleration during contact with the ground
is

Dv 0 - (-17.1 m/s)
aavg = = = 857 m/s 2 .
Dt 20.0 ¥ 10-3 s

(b) The fact that the result is positive indicates that this acceleration vector points
upward.

LEARN Since D t is very small, it is not surprising to have a very large acceleration
to stop the motion of the ball. In later chapters, we shall see that the acceleration is
directly related to the magnitude and direction of the force exerted by the ground on
the ball during the course of collision.
80 CHAPTER 2

56. We use Eq. 2-16,

vB2 = vA2 + 2a(yB yA),
1
with a = 9.8 m/s2, yB yA = 0.40 m, and vB = 3
vA. It is then straightforward to solve:
vA = 3.0 m/s, approximately.
 81

57. The average acceleration during contact with the floor is aavg = (v2 v1) / Dt,
where v1 is its velocity just before striking the floor, v2 is its velocity just as it leaves
the floor, and Dt is the duration of contact with the floor (12 ¥ 10 3 s).

(a) Taking the y axis to be positively upward and placing the origin at the point where
the ball is dropped, we first find the velocity just before striking the floor, using
v12 = v02 - 2 gy . With v0 = 0 and y = 4.00 m, the result is

v1 = - -2 gy = - -2(9.8 m/s 2 ) (-4.00 m) = -8.85 m/s

where the negative root is chosen because the ball is traveling downward. To find the
velocity just after hitting the floor (as it ascends without air friction to a height of 2.00
m), we use v 2 = v22 - 2 g ( y - y0 ) with v = 0, y = 2.00 m (it ends up two meters
below its initial drop height), and y0 = 4.00 m. Therefore,

v2 = 2 g ( y - y0 ) = 2(9.8 m/s 2 ) (-2.00 m + 4.00 m) = 6.26 m/s .

Consequently, the average acceleration is

v2 - v1 6.26 m/s - (- 8.85 m/s)

aavg = = -3
= 1.26 ¥ 103 m/s 2 .
Dt 12.0 ¥ 10 s

(b) The positive nature of the result indicates that the acceleration vector points
upward. In a later chapter, this will be directly related to the magnitude and direction
of the force exerted by the ground on the ball during the collision.
82 CHAPTER 2

58. We choose down as the +y direction and set the coordinate origin at the point
where it was dropped (which is when we start the clock). We denote the 1.00 s
duration mentioned in the problem as t t' where t is the value of time when it lands
and t' is one second prior to that. The corresponding distance is y y' = 0.60h, where y
denotes the location of the ground. In these terms, y is the same as h, so we have h y'
= 0.60h or 0.40h = y' .

(a) We find t' and t from Eq. 2-15 (with v0 = 0):

1 2 y¢
y¢ = gt ¢2 fi t¢ =
2 g
1 2 2y
y= gt fi t = .
2 g

Plugging in y = h and y' = 0.40h, and dividing these two equations, we obtain

t¢ 2 ( 0.40h ) / g
= = 0.40 .
t 2h / g

Letting t' = t 1.00 (SI units understood) and cross-multiplying, we find

1.00
t - 1.00 = t 0.40 fi t =
1 - 0.40
which yields t = 2.72 s.

(b) Plugging this result into y = 12 gt 2 we find h = 36 m.

(c) In our approach, we did not use the quadratic formula, but we did choose a root
when we assumed (in the last calculation in part (a)) that 0.40 = +0.632
instead of 0.707. If we had instead let 0.40 = 0.632 then our answer for t
would have been roughly 0.6 s, which would imply that t' = t 1 would equal a
negative number (indicating a time before it was dropped), which certainly does
not fit with the physical situation described in the problem.
.
 83

59. We neglect air resistance, which justifies setting a = g = 9.8 m/s2 (taking down
as the y direction) for the duration of the motion. We are allowed to use Table 2-1
(with Dy replacing Dx) because this is constant acceleration motion. The ground level
is taken to correspond to the origin of the y-axis.

(a) The time drop 1 leaves the nozzle is taken as t = 0 and its time of landing on the
floor t1 can be computed from Eq. 2-15, with v0 = 0 and y1 = 2.00 m.

1 -2 y -2( -2.00 m)
y1 = - gt12 fi t1 = = = 0.639 s .
2 g 9.8 m/s 2

At that moment, the fourth drop begins to fall, and from the regularity of the
dripping we conclude that drop 2 leaves the nozzle at t = 0.639/3 = 0.213 s and
drop 3 leaves the nozzle at t = 2(0.213 s) = 0.426 s. Therefore, the time in free
fall (up to the moment drop 1 lands) for drop 2 is t2 = t1 0.213 s = 0.426 s. Its
position at the moment drop 1 strikes the floor is

1 1
y2 = - gt22 = - (9.8 m/s 2 )(0.426 s) 2 = -0.889 m,
2 2

or about 89 cm below the nozzle.

(b) The time in free fall (up to the moment drop 1 lands) for drop 3 is t3 = t1 0.426 s
= 0.213 s. Its position at the moment drop 1 strikes the floor is

1 1
y3 = - gt32 = - (9.8 m/s 2 )(0.213 s) 2 = -0.222 m,
2 2

or about 22 cm below the nozzle.

84 CHAPTER 2

60. To find the launch velocity of the rock, we apply Eq. 2-11 to the maximum
height (where the speed is momentarily zero)

v = v0 - gt fi 0 = v0 - ( 9.8 m/s 2 ) ( 2.5 s )

so that v0 = 24.5 m/s (with +y up). Now we use Eq. 2-15 to find the height of the
tower (taking y0 = 0 at the ground level)

1 2 1
at fi y - 0 = ( 24.5 m/s )(1.5 s ) - ( 9.8 m/s 2 ) (1.5 s ) .
2
y - y0 = v0t +
2 2

Thus, we obtain y = 26 m.
 85

61. We choose down as the +y direction and place the coordinate origin at the top of
the building (which has height H). During its fall, the ball passes (with velocity v1) the
top of the window (which is at y1) at time t1, and passes the bottom (which is at y2) at
time t2. We are told y2 y1 = 1.20 m and t2 t1 = 0.125 s. Using Eq. 2-15 we have

b
y2 - y1 = v1 t2 - t1 + g 1
2
b
g t 2 - t1 g 2

which yields

v1 =
1.20 m - 1
2 ( 9.8 m/s ) ( 0.125 s )
2 2

= 8.99 m/s.
0.125 s

From this, Eq. 2-16 (with v0 = 0) reveals the value of y1:

(8.99 m/s)2
v12 = 2 gy1 fi y1 = = 4.12 m.
2(9.8 m/s 2 )

It reaches the ground (y3 = H) at t3. Because of the symmetry expressed in the
problem ( upward flight is a reverse of the fall ) we know that t3 t2 = 2.00/2 = 1.00
s. And this means t3 t1 = 1.00 s + 0.125 s = 1.125 s. Now Eq. 2-15 produces
1
y3 - y1 = v1 (t3 - t1 ) + g (t3 - t1 )2
2
1
y3 - 4.12 m = (8.99 m/s) (1.125 s) + (9.8 m/s 2 ) (1.125 s) 2
2

which yields y3 = H = 20.4 m.

86 CHAPTER 2

62. The height reached by the player is y = 0.78 m (where we have taken the origin of
the y axis at the floor and +y to be upward).

(a) The initial velocity v0 of the player is

v0 = 2 gy = 2(9.8 m/s 2 ) (0.78 m) = 3.91 m/s .

This is a consequence of Eq. 2-16 where velocity v vanishes. As the player reaches y1
= 0.78 m 0.15 m = 0.63 m, his speed v1 satisfies v02 - v12 = 2 gy1 , which yields

v1 = v02 - 2 gy1 = (3.91 m/s)2 - 2(9.80 m/s 2 ) (0.63 m) = 1.71 m/s .

The time t1 that the player spends ascending in the top Dy1 = 0.15 m of the jump can
now be found from Eq. 2-17:
1 2 ( 0.15 m )
Dy1 = ( v1 + v ) t1 fi t1 = = 0.175 s
2 1.71 m/s + 0

which means that the total time spent in that top 15 cm (both ascending and
descending) is 2(0.175 s) = 0.35 s = 350 ms.

(b) The time t2 when the player reaches a height of 0.15 m is found from Eq. 2-15:

1 2 1
0.15 m = v0 t2 - gt2 = (3.91 m/s)t2 - (9.8 m/s 2 )t22 ,
2 2

which yields (using the quadratic formula, taking the smaller of the two positive roots)
t2 = 0.0404 s = 40.4 ms, which implies that the total time spent in that bottom 15 cm
(both ascending and descending) is 2(40.4 ms) = 80.8 ms, or about 81 ms.
 87

63. The time t the pot spends passing in front of the window of length L = 2.0 m is
0.25 s each way. We use v for its velocity as it passes the top of the window (going
up). Then, with a = g = 9.8 m/s2 (taking down to be the y direction), Eq. 2-18
yields
1 L 1
L = vt - gt 2 fi v = - gt .
2 t 2

The distance H the pot goes above the top of the window is therefore (using Eq. 2-16
with the final velocity being zero to indicate the highest point)

( 2.00 m / 0.25 s - (9.80 m/s 2 )(0.25 s) / 2 )

2
v 2 ( L / t - gt / 2 )
2

H= = = = 2.34 m.
2g 2g 2(9.80 m/s 2 )
88 CHAPTER 2

64. Using the equation of motion

1
( x - x0 ) = v0 t + at 2 ,
2
we get the rock s displacement during the fall as follows:
1 2
h = ( x - x0 ) = v0 t + at
2
È1 ˘
= -10(3) + Í ( -9.8)(3) 2 ˙ (acceleration here is the acceleration due to gravity, - 9.8 m/s 2 )
Î 2 ˚
= -74 m.

Thus the initial height is 74 m.

 89

65. The key idea here is that the speed of the head (and the torso as well) at any given
time can be calculated by finding the area on the graph of the head s acceleration
versus time, as shown in Eq. 2-26:

Ê area between the acceleration curve ˆ

v1 - v0 = Á ˜
Ë and the time axis, from t0 to t1 ¯

(a) From Fig. 2.15a, we see that the head begins to accelerate from rest (v0 = 0) at t0 =
110 ms and reaches a maximum value of 90 m/s2 at t1 = 160 ms. The area of this
region is
1
area = (160 - 110) ¥10 -3s 6 ( 90 m/s 2 ) = 2.25 m/s
2

which is equal to v1, the speed at t1.

(b) To compute the speed of the torso at t1=160 ms, we divide the area into 4 regions:
From 0 to 40 ms, region A has zero area. From 40 ms to 100 ms, region B has the
shape of a triangle with area
1
area B = (0.0600 s)(50.0 m/s 2 ) = 1.50 m/s .
2
From 100 to 120 ms, region C has the shape of a rectangle with area

area C = (0.0200 s) (50.0 m/s 2 ) = 1.00 m/s.

From 110 to 160 ms, region D has the shape of a trapezoid with area
1
area D = (0.0400 s) (50.0 + 20.0) m/s 2 = 1.40 m/s.
2

Substituting these values into Eq. 2-26, with v0 = 0 then gives

v1 - 0 = 0 + 1.50 m/s + 1.00 m/s + 1.40 m/s = 3.90 m/s,

or v1 = 3.90 m/s.
90 CHAPTER 2

66. The key idea here is that the position of an object at any given time can be
calculated by finding the area on the graph of the object s velocity versus time, as
shown in Eq. 2-30:
Ê area between the velocity curve ˆ
x1 - x0 = Á ˜.
Ë and the time axis, from t0 to t1 ¯

(a) To compute the position of the fist at t = 50 ms, we divide the area in Fig. 2-37
into two regions. From 0 to 10 ms, region A has the shape of a triangle with area
1
area A = (0.010 s) (2 m/s) = 0.01 m.
2

From 10 to 50 ms, region B has the shape of a trapezoid with area

1
area B = (0.040 s) (2 + 4) m/s = 0.12 m.
2

Substituting these values into Eq. 2-30 with x0 = 0 then gives

x1 - 0 = 0 + 0.01 m + 0.12 m = 0.13 m,
or x1 = 0.13 m.

(b) The speed of the fist reaches a maximum at t1 = 120 ms. From 50 to 90 ms, region
C has the shape of a trapezoid with area
1
area C = (0.040 s) (4 + 5) m/s = 0.18 m.
2

From 90 to 120 ms, region D has the shape of a trapezoid with area
1
area D = (0.030 s) (5 + 7.5) m/s = 0.19 m.
2

Substituting these values into Eq. 2-30, with x0 = 0 then gives

x1 - 0 = 0 + 0.01 m + 0.12 m + 0.18 m + 0.19 m = 0.50 m,

or x1 = 0.50 m.
 91

67. The problem is solved using Eq. 2-31:

Ê area between the acceleration curve ˆ

v1 - v0 = Á ˜
Ë and the time axis, from t0 to t1 ¯

To compute the speed of the unhelmeted, bare head at t1 = 7.0 ms, we divide the area
under the a vs. t graph into 4 regions: From 0 to 2 ms, region A has the shape of a
triangle with area
1
area A = (0.0020 s) (120 m/s 2 ) = 0.12 m/s.
2
From 2 ms to 4 ms, region B has the shape of a trapezoid with area

1
area B = (0.0020 s) (120 + 140) m/s 2 = 0.26 m/s.
2

From 4 to 6 ms, region C has the shape of a trapezoid with area

1
area C = (0.0020 s) (140 + 200) m/s 2 = 0.34 m/s.
2

From 6 to 7 ms, region D has the shape of a triangle with area

1
area D = (0.0010 s) (200 m/s 2 ) = 0.10 m/s.
2

Substituting these values into Eq. 2-31, with v0=0 then gives

vunhelmeted = 0.12 m/s + 0.26 m/s + 0.34 m/s + 0.10 m/s = 0.82 m/s.

Carrying out similar calculations for the helmeted head, we have the following
results: From 0 to 3 ms, region A has the shape of a triangle with area
1
area A = (0.0030 s) (40 m/s 2 ) = 0.060 m/s.
2
From 3 ms to 4 ms, region B has the shape of a rectangle with area

area B = (0.0010 s) (40 m/s 2 ) = 0.040 m/s.

From 4 to 6 ms, region C has the shape of a trapezoid with area

1
area C = (0.0020 s) (40 + 80) m/s 2 = 0.12 m/s.
2
From 6 to 7 ms, region D has the shape of a triangle with area
1
area D = (0.0010 s) (80 m/s 2 ) = 0.040 m/s.
2
92 CHAPTER 2

Substituting these values into Eq. 2-31, with v0 = 0 then gives

vhelmeted = 0.060 m/s + 0.040 m/s + 0.12 m/s + 0.040 m/s = 0.26 m/s.

Thus, the difference in the speed is

Dv = vunhelmeted - vhelmeted = 0.82 m/s - 0.26 m/s = 0.56 m/s.

 93

68. This problem can be solved by noting that velocity can be determined by the
graphical integration of acceleration versus time. The speed of the tongue of the
salamander is simply equal to the area under the acceleration curve:

1 -2 1 1
v = area = (10 s)(100 m/s 2 ) + (10-2 s)(100 m/s 2 + 400 m/s 2 ) + (10 -2 s)(400 m/s 2 )
2 2 2
= 5.0 m/s.
94 CHAPTER 2

z
69. Since v = dx / dt (Eq. 2-4), then Dx = v dt , which corresponds to the area
under the v vs t graph. Dividing the total area A into rectangular (base ¥ height) and
b g
triangular 21 base ¥ height areas, we have

A = A0 <t < 2 + A2 <t <10 + A10 < t <12 + A12 < t <16

=
1 FG 1
(2)(8) + (8)(8) + (2)(4) + (2 )(4) + (4 )(4)
IJ
2 H 2 K
with SI units understood. In this way, we obtain Dx = 100 m.
 95

70. To solve this problem, we note that velocity is equal to the time derivative of a
position function, as well as the time integral of an acceleration function, with the
integration constant being the initial velocity. Thus, the velocity of particle 1 can be
written as
dx d
v1 = 1 = ( 6.00t 2 + 3.00t + 2.00 ) = 12.0t + 3.00 .
dt dt

Similarly, the velocity of particle 2 is

v2 = v20 + Ú a2 dt = 15.0 + Ú (-8.00t )dt = 20.0 - 4.00t 2 .

The condition that v1 = v2 implies

12.0t + 3.00 = 15.0 - 4.00t 2 fi 4.00t 2 + 12.0t - 12.0 = 0

which can be solved with the quadratic equation to give t = 0.790 s. Thus, the
velocity at this time is v1 = v2 = 12.0(0.790) + 3.00 = 12.5 m/s.