21
Chapter 2
1. We use Eq. 2-2 and Eq. 2-3. During a time
t
c
when the velocity remains a positive
constant, speed is equivalent to velocity, and distance is equivalent to displacement,
with
Δ
x
=
v t
c
.
(a) During the first part of the motion, the displacement is
Δ
x
1
= 40 km and the time
interval is
t
1
40
133
=
=
(
.
km)
(30 km / h)
h.
During the second part the displacement is
Δ
x
2
= 40 km and the time interval is
t
2
40
0 67
=
=
(
.
km)
(60 km / h)
h.
Both displacements are in the same direction, so the total displacement is
Δ
x
=
Δ
x
1
+
Δ
x
2
= 40 km + 40 km = 80 km.
The total time for the trip is
t
=
t
1
+
t
2
= 2.00 h. Consequently, the average velocity is
v
avg
km)
(2.0 h)
km / h.
=
=
(80
40
(b) In this example, the numerical result for the average speed is the same as the
average velocity 40 km/h.
(c) As shown below, the graph consists of two contiguous line segments, the first
having a slope of 30 km/h and connecting the origin to (
t
1
,
x
1
) = (1.33 h, 40 km) and
the second having a slope of 60 km/h and connecting (
t
1
,
x
1
) to (
t, x
) = (2.00 h, 80 km).
From the graphical point of view, the slope of the dashed line drawn from the origin
to (
t, x
) represents the average velocity.

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CHAPTER 2
22
2. 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
D
D
t
t
D
D
v
D
v
up
down
up
down
up
down
+
+
=
+
2
which, after canceling
D
and plugging in
v
up
= 40 km/h and
v
down
= 60 km/h, yields 48
km/h for the average speed.
3. The speed (assumed constant) is
v
= (90 km/h)(1000 m/km)
⁄
(3600 s/h) = 25 m/s.
Thus, in 0.50 s, the car travels (0.50 s)(25 m/s)
≈
13 m.
4. Huber’s speed is
v
0
= (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
v
1
=(110.6 km/h + 19.0 km/h)=129.6 km/h, or 36 m/s
(1 km/h = 0.2778 m/s). Thus, the time through a distance of 200 m for Whittingham is
1
200 m
5.554 s.
36 m/s
x
t
v
Δ
Δ =
=
=
5. Using
x
= 3
t
– 4
t
2
+
t
3
with SI units understood is efficient (and is the approach we
will use), but if we wished to make the units explicit we would write
x
= (3 m/s)
t
– (4 m/s
2
)
t
2
+ (1 m/s
3
)
t
3
.
We will quote our answers to one or two significant figures, and not try to follow the
significant figure rules rigorously.
(a) Plugging in
t
= 1 s yields
x
= 3 – 4 + 1 = 0.
(b) With
t
= 2 s we get
x
= 3(2) – 4(2)
2
+(2)
3
= –2 m.
(c) With
t
= 3 s we have
x
= 0 m.
(d) Plugging in
t
= 4 s gives
x
= 12 m.
For later reference, we also note that the position at
t
= 0 is
x
= 0.
(e) The position at
t
= 0 is subtracted from the position at
t
= 4 s to find the
displacement
Δ
x
= 12 m.
(f) The position at
t
= 2 s is subtracted from the position at
t
= 4 s to give the
displacement
Δ
x
= 14 m. Eq. 2-2, then, leads to