2.1: a) During the later 4.75-s interval, the rocket moves a distance
m63m1000.1
3

, and so the magnitude of the average velocity is
.sm197
s754
m63m1000.1
3


.
b)
sm169
s5.90
m1000.1
3


2.2: a) The magnitude of the average velocity on the return flight is
.sm42.4
)das400,86()da5.13(
)m105150(
3


The direction has been defined to be the –x-direction
).
ˆ
( i
b) Because the bird ends up at the starting point, the average velocity for the round

).
ˆ
( i
2.5: In time t the fast runner has traveled 200 m farther than the slow runner:
s286so,s)m(6.20m200s)m50.5(  ttt
.
Fast runner has run
m.1770)sm20.6( t
Slow runner has run
m.1570)sm50.5( t
2.6: The s-waves travel slower, so they arrive 33 s after the p-waves.
km250
s33
5.65.3
s33
s33
s
km
s
km
ps
ps





d
dd
v

s4.00
0m8.20


c)
sm6.7
s2.00
m5.60m8.20


2.9: a) At
0 ,0
11
 xt
, so Eq (2.2) gives
.sm0.12
s)0.10(
s)0.10)(sm120.0(s)0.10)(sm4.2(
3
3
22
2
2
av



t
x
v

. Therefore
0)sm360.0()sm80.4(
2
32
 tt
. The
only time after
0t
when the car is at rest is
s3.13
3
2
sm360.0
sm80.4
t
2.10: a) IV: The curve is horizontal; this corresponds to the time when she stops. b) I:
This is the time when the curve is most nearly straight and tilted upward (indicating
postive velocity). c) V: Here the curve is plainly straight, tilted downward (negative
velocity). d) II: The curve has a postive slope that is increasing. e) III: The curve is still
tilted upward (positive slope and positive velocity), but becoming less so.
2.11: Time (s) 0 2 4 6 8 10 12 14
16
Acceleration (m/s
2
) 0 1 2 2 3 1.5 1.5 0
a) The acceleration is not constant, but is approximately constant between the times
s4t
and
s.8t
2.12: The cruising speed of the car is 60

ˆ
)sm0.3()sm0.7(
ˆ
)sm0.10(
ˆ
)sm0.5()(
22
ttttt 

The velocity vector is the time derivative of the displacement vector:
kji
r
ˆ
))sm0.3(2sm0.7(
ˆ
)sm0.10(
ˆ
)sm0.5(
)(
2
t
dt
td


and the acceleration vector is the time derivative of the velocity vector:
k
r
ˆ
sm0.6


kji
kji
r
ˆ
)sm0.23(
ˆ
)sm0.10(
ˆ
)sm0.5(
ˆ
))s0.5)(sm0.6(sm0.7((
ˆ
)sm0.10(
ˆ
)sm0.5(
)(
2


dt
td

k
r
ˆ
sm0.6
)(
2
2

avxt
b)
s.0.16:for solveand0Set  ttv
x
c) Set x = 50.0 cm and solve for t. This gives
s.0.32 and0  tt
The turtle returns
to the starting point after 32.0 s.
d) Turtle is 10.0 cm from starting point when x = 60.0 cm or x = 40.0 cm.
s.8.25ands20.6:for solveandcm0.60Set  tttx
.scm23.1s,8.25At
.scm23.1s,20.6At


x
x
vt
vt
Set
cm0.40x
and solve for
s4.36: tt
(other root to the quadratic equation is negative
and hence nonphysical).
.scm55.2 s,4.36At 
x
vt
e)
2.16: Use of Eq. (2.5), with t = 10 s in all cases,
a)

,
and the velocity at t = 5.00 s is
(3.00
sm
) + (0.100
3
sm
) (5.00 s)
2
= 5.50
sm
,
so Eq. (2.4) gives the average acceleration as
2
sm50.
)s00.5(
)sm00.3()sm50.5(


.
b) The instantaneous acceleration is obtained by using Eq. (2.5),
.)sm2.0(2
3
tt
dt
dv
a
x



dt
dv
a
x

There are two times at which v = 0 (three if negative times are considered), given by t =
0 and t
4
= 16 s
4
. At t = 0, x = 2.17 m and a
x
= 9.60
sm
2
. When t
4
= 16 s
4
,
x = (2.17 m) + (4.80
sm
2
)
)s16(
4
– (0.100)
6
sm
)(16 s


.
v
t
xx
v
xx
b) The above result for v
0x
may be used to find
,sm43.1
s00.7
sm00.5sm0.15
2
0





t
vv
a
xx
x
or the intermediate calculation can be avoided by combining Eqs. (2.8) and (2.12) to
eliminate v
0x
and solving for a
x

t
v
a
x
x
2.22: a) The acceleration is found from Eq. (2.13), which
x
v
0
= 0;
  
  
,sm0.32
)ft307(2
)hrmi173(
)(2
2
ft3.281
m1
2
hrmi1
sm4470.0
0
2



xx
v
a

0



x
v
xx
t
2.23: From Eq. (2.13), with
 
,0 Taking.,0
0max
2
0
2
0


xaav
xx
v
xx
m.70.1
)sm250(2
))hrkms)(3.6mhr)(1km105((
2
2
2
max
2

)sm20(
)(2
2
2
0
2



xx
v
a
x
x
b) Using Eq. (2.14),
s.12)sm20(m)120(2)(2
0
 vxxt
c)
m.240)sm20)(s12( 
2.26: a) x
0
< 0, v
0x
< 0, a
x
< 0
b) x
0
> 0, v

.sft5.26gives)(2
2
0
2
0
2

xxxx
axxavv
b)
? ,sft676,0ft,1320
2
00

xxx
v.avxx
.mph5.90sft133gives)(2
0
2
0
2

xxxx
vxxavv
constant.benot must
x
a
c)
?,0,sft5.26 s,ft0.88
2

6s5.4
2
1
s
cm
2s5.4















From 0 to 7.5 s:
The distance is the sum of the magnitudes of the areas.
   
cm5.25
s
cm
2s5.1
2
1

sm3.6

. From
t = 9 s to t = 13 s the acceleration is constant and equal to
.sm2.11
2
s4
sm450


b) In the first five seconds, the area under the graph is the area of the rectangle, (20
m)(5 s) = 100 m. Between t = 5 s and t = 9 s, the area under the trapezoid is (1/2)(45 m/s
+ 20 m/s)(4 s) = 130 m (compare to Eq. (2.14)), and so the total distance in the first 9 s is
230 m. Between t = 9 s and t = 13 s, the area under the triangle is
m90s)4)(sm45)(21( 
, and so the total distance in the first 13 s is 320 m.
2.32:
2.33: a) The maximum speed will be that after the first 10 min (or 600 s), at which time
the speed will be
.skm18sm101.8s)900)(sm0.20(
4
2

b) During the first 15 minutes (and also during the last 15 minutes), the ship will travel
km8100s)900)(skm18)(21( 
, so the distance traveled at non-constant speed is 16,200
km and the fraction of the distance traveled at constant speed is
,958.0
km384,000
km200,16

2

During the 70-second period when the train moves with constant speed, the train travels
  
m.1568s70sm4.22 
The distance traveled during deceleration is given by Eq.
(2.13), with
sm4.22,0
0

xx
vv
and
2
sm50.3
x
a
, so the train moves a distance
.m68.71
)m/s3.502(
)s/m4.22(
0
2
2



xx
The total distance covered in then 156.8 m + 1568 m + 71.7 m
= 1.8 km.

11
211
2
11T
















a
ta
tt
ta
a
ta
ttatax
2.35: a)
b) From the graph (Fig. (2.35)), the curves for A and B intersect at t = 1 s and t = 3 s.
c)

t
and at this time
x
T
= x
C
= 250 m.
b) a
C
t = (3.20 m/s
2
)(12.5 s) = 40.0 m/s (See Exercise 2.37 for a discussion of why the
car’s speed at this time is twice the truck’s speed.)
c)
d)
2.37: a)
The car and the motorcycle have gone the same distance during the same time, so their
average speeds are the same. The car's average speed is its constant speed v
C
, and for
constant acceleration from rest, the motorcycle's speed is always twice its
average, or 2v
C
. b) From the above, the motorcyle's speed will be v
C
after half the time
needed to catch the car. For motion from rest with constant acceleration, the distance
traveled is proportional to the square of the time, so for half the time
one-fourth of the total distance has been covered, or
.4d

)m/s(9.80
)m440.0(2
2
)(2
2
)(2
2
2
0
0





g
yy
g
yyg
t
or about 0.60 s.
2.40: Using Eq. (2.13), with downward velocities and accelerations being positive,
2
y
v
=
(0.8
sm
)
2



2.42: a)
m.6.30s)5.2)(sm80.9)(21()21(
222
gt
b)
sm.5.24s)5.2)(sm80.9(
2
gt
.
c)
2.43: a) Using the method of Example 2.8, the time the ring is in the air is
)sm80.9(
)m0.12()sm80.9(2s)m00.5(s)m00.5(
)(2
2
2
2
0
2
00




g
yygvv
t
yy

2
00
2
1
gttvyy
y

22
)sm8.9(
2
1
)sm(5.00m0.120
tt 
Solve this quadratic as in part a) to obtain t = 2.156 s.
d)
)m0.12)(sm8.9(2)sm(5.00)(2
22
0
2
0
2
 yygvv
yy

sm1.16
y
v
e)
2.44: a) Using a
y

= 40.1 m,
v
y
= (5.00
sm
) – (9.80
2
sm
)(1.00 s) = – 4.80
sm
.
b) Using the result derived in Example 2.8, the time is
t =
)sm80.9(
)m0.400)(sm80.9(2)sm00.5()sm00.5(
2
2
2

= 3.41 s.
c) Either using the above time in Eq. (2.8) or avoiding the intermediate calculation by
using Eq. (2.13),
,sm809)m0.40)(sm80.9(2)sm00.5()(2
2
2
2
2
0
2
0

yy
so the speed is
sm6.25
.
b)
m,6.31s)00.2)(sm80.9(
2
1
s)00.2)(sm00.6(
2
1
2
2
2
0
 gttvy
y
with the
minus sign indicating that the balloon has indeed fallen.
c)
m2.15so,sm232m)0.10)(sm80.9(2s)m00.6()(2
2222
0
2
0
2

yyy
vyygvv
2.46: a) The vertical distance from the initial position is given by

0
2
0
2
yygvv
yy

with v = 0, to solve for y
– y
0
= 10.7 m (this requires retention of two extra significant figures in the calculation
for v
0y
). c) 0 d) 9.8
2
sm
, down.
e) Assume the top of the building is 50 m above the ground for purposes of graphing:
2.47: a)
.sm249s)9.0()sm224(
2

b)
.4.25
2
2
sm9.80
sm249

c) The most direct way to

2


(This ignores the t = 0 solution.)
d) Again from Eq. (2.8), (40
sm
)/(9.80
2
sm
) = 4.08 s. This is, of course, half the
time found in part (c).
e) 9.80
2
sm
, down, in all cases.
f)
2.49: a) For a given initial upward speed, the height would be inversely proportional to
the magnitude of g, and with g one-tenth as large, the height would be ten times higher,
or 7.5 m. b) Similarly, if the ball is thrown with the same upward speed, it would go ten
times as high, or 180 m. c) The maximum height is determined by the speed when hitting
the ground; if this speed is to be the same, the maximum height would be ten times as
large, or 20 m.
2.50: a) From Eq. (2.15), the velocity v
2
at time t



t
t

sm
) + (0.6
3
sm
) t
2
.
At t
2
= 2.0 s, the velocity is v
2
= (4.40
sm
) + (0.6
3
sm
)(2.0 s)
2
= 6.80 m/s, or 6.8
sm
to two significant figures.
b) From Eq. (2.16), the position x
2
as a function of time is

dtvxx
x
t
t
1

3
)
= 11.8 m.
c)