CHAPTER
2
Motion in One Dimension
1* ·
What is the approximate average velocity of the race cars during the Indianapolis 500?
Since the cars go around a closed circuit and return nearly to the starting point, the displacement is nearly zero,
and the average velocity is zero.
2
·
Does the following statement make sense? “The average velocity of the car at 9 a.m. was 60 km/h.”
No, it does not. Average velocity must refer to a finite time interval.
3
·
Is it possible for the average velocity of an object to be zero during some interval even though its average
velocity for the first half of the interval is not zero? Explain.
Yes, it is. In a round trip,
A
to
B
and back to
A
, the average velocity is zero; the average velocity between
A
and
B
is not zero.
4
·
The diagram in Figure 2-21 tracks the path of an object moving in a straight line. At which point is the object
farthest from its starting point?
(
a
)
A
(
b
)
B
(
c
)
C
(
d
)
D
(
e
)
E
(
b
) Starting point is at
x
= 0; point
B
is farthest from
x
= 0.
5* ·
(
a
) An electron in a television tube travels the 16-cm distance from the grid to the screen at an average
speed of 4
×
10
7
m/s. How long does the trip take?
(
b
) An electron in a current-carrying wire travels at an
average speed of 4
×
10
-5
m/s. How long does it take to travel 16 cm?
(
a
) From Equ. 2-3,
∆
t
=
∆
s
/(av. speed)
(
b
) Repeat as in (
a
)
∆
t
= (0.16 m)/(4
×
10
7
m/s) = 4
×
10
-9
s = 4 ns
∆
t
= (0.16 m)/(4
×
10
-5
m/s) = 4
×
10
3
s = 4 ks
6
·
A runner runs 2.5 km in 9 min and then takes 30 min to walk back to the starting point.
(
a
) What is the
runner’s average velocity for the first 9 min?
(
b
) What is the average velocity for the time spent walking?
(
c
) What
is the average velocity for the whole trip?
(
d
) What is the average speed for the whole trip?
Take the direction of running as the positive direction.
(
a
) Use Equ. 2-2
(
b
) Use Equ. 2-2
(
c
)
∆
x
= 0
(
d
) Total distance = 5.0 km;
∆
t
= 39 min
v
av
= (2.5 km)/[(9 min)(1 h/60 min)] = 16.7 km/h
v
av
= (-2.5 km)/(0.5 h) = -5.0 km/h
v
av
= 0
Av. speed = (5.0 km)/[(39 min)(1 h/60 min)]
= 7.7 km/h

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