Chapter 2
Page 1
CHAPTER 2 - Describing Motion: Kinematics in One Dimension
1.
We find the time from
average speed =
d
/
t
;
15 km/h = (75 km)/
t
, which gives
t
= 5.0 h
.
2.
We find the average speed from
average speed =
d
/
t
= (280 km)/(3.2 h) =
88 km/h
.
3.
We find the distance traveled from
average speed =
d
/
t
;
(110 km/h)/(3600 s/h) =
d
/(2.0 s), which gives
d
= 6.1
×
10
–2
km = 61 m
.
4.
We find the average velocity from
æ
= (
x
2
–
x
1
)/(
t
2
–
t
1
) = (– 4.2 cm – 3.4 cm)/(6.1 s – 3.0 s) =
– 2.5 cm/s (toward –
x
)
.
5.
We find the average velocity from
æ
= (
x
2
–
x
1
)/(
t
2
–
t
1
) = (8.5 cm – 3.4 cm)/[4.5 s – (– 2.0 s)] =
0.78 cm/s (toward +
x
)
.
Because we do not know the total distance traveled, we cannot calculate the average speed.
6.
(
a
)
We find the elapsed time before the speed change from
speed =
d
1
/
t
1
;
65 mi/h = (130 mi)/
t
1
, which gives
t
1
= 2.0 h.
Thus the time at the lower speed is
t
2
=
T
–
t
1
= 3.33 h – 2.0 h = 1.33 h.
We find the distance traveled at the lower speed from
speed =
d
2
/
t
2
;
55 mi/h =
d
2
/(1.33 h), which gives
d
2
= 73 mi.
The total distance traveled is
D
=
d
1
+
d
2
= 130 mi + 73 mi =
203 mi
.
(
b
)
We find the average speed from
average speed =
d
/
t
= (203 mi)/(3.33 h) =
61 mi/h
.
Note that the average speed is
not
!
(65 mi/h + 55 mi/h).
The two speeds were not maintained for
equal times.
7.
Because there is no elapsed time when the light arrives, the sound travels one mile in 5 seconds.
We find the speed of sound from
speed =
d
/
t
= (1 mi)(1610 m/1 mi)/(5 s)
≈
300 m/s
.
8.
(
a
)
We find the average speed from
average speed =
d
/
t
= 8(0.25 mi)(1.61
×
10
3
m/mi)/(12.5 min)(60 s/min) =
4.29 m/s
.
(
b
)
Because the person finishes at the starting point, there is no displacement;
thus the average velocity is
æ
=
∆
x
/
∆
t
=
0
.
9.
(
a
)
We find the average speed from
average speed =
d
/
t
= (160 m + 80 m)/(17.0 s + 6.8 s) =
10.1 m/s
.
(
b
)
The displacement away from the trainer is 160 m – 80 m = 80 m; thus the average velocity is
æ
=
∆
x
/
∆
t
= (80 m)/(17.0 s + 6.8 s) =
+ 3.4 m/s, away from trainer
.
10.
Because the two locomotives are traveling with equal speeds in opposite directions, each locomotive will travel half
the distance, 4.25 km.
We find the elapsed time from
speed
1
=
d
1
/
t
;
(95 km/h)/(60 min/h) = (4.25 km)/
t
, which gives
t
= 2.7 min
.

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