This preview shows pages 1–3. Sign up to view the full content.
1
1
Solutions to Assigned Problems, Unit 2, Chapter 4
MOTION WITH A CHANGING VELOCITY
1. (a) Strategy
Relate the acceleration, speed, and distance using Eq. (45).
Solution
22
2
2
fi
0(
3
.
2
m
s
)
2
, so
0.85 m s .
2
2(6.0 m
0)
xx
x
x
vv
a
x
a
x
−
−
−=
Δ
=
=
=
−
Δ−
The magnitude of the average acceleration is
2
0.85 m s
.
(b) Strategy
Use Newton’s second law. Draw a freebody diagram.
S
o
l
u
t
i
o
n
Find
k
.
μ
kk
k
k
or
.
x
x
ma
F
f
ma
f
N
ma
N
μμ
∑=
=
=
−
=
−
0, so
.
y
F
Nm
g
N m
g
−
=
=
Thus, we have
2
k
2
0.85 m s
0.087 .
9.80 m s
x
ma
ma
a
g
g
−
=−
=
f
N
m
g
y
x
v
k
5. (a)
Strategy
Use Eqs. (33) and (42) since the acceleration is constant.
Solution
Find the distance traveled.
av,
27.3 m s 17.4 m s
(10.0 s)
224 m .
x
xv
t
t
+
+
Δ=
=
(b)
Strategy
Use the definition of average acceleration.
Solution
Find the magnitude of the acceleration.
2
27
.3
ms 17
.4
ms
0.99 m s
10.0 s
v
a
t
==
=
Δ
9. Strategy
Use Eq. (45) to find the distance the car requires to stop after slamming on the brakes. Then, use Eqs.
(33) and (42) to relate the speeds, distances, time, and acceleration.
Solution
Find the distance the car requires to stop.
2
fc
ic
c
2
2
7
.
0
m
s
)
52.1 m .
2
2( 7.00 m s )
x
a
−
−
=
=
−
Since the acceleration of the car is constant, the average speed of
the car as it attempts to stop is
c,av
fc
ic
(
)2 (0 27
.0
)2 13
.5
.
v
=
+=
+
=
Thus, the time required for the
car to stop is
cc
,
a
v
(52.1 m) (13.5 m s)
3.86 s.
tx
v
Δ=Δ
=
=
The distance the tractor travels in this time is
tt
(10.0 m s)(3.86 s)
38.6 m.
t
Δ=Δ=
=
Now, 38.6 m 15.0 m
53.6 m,
+
=
which is
53.6 m
52.1 m
1.5 m
beyond the stopping point of the car. Therefore, you won’t hit the tractor.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentChapter 4:
Motion with a Changing Velocity
College Physics
2
13. (a)
Strategy
The graph will be a line with a slope of
2
1.20 m s .
Solution
0 when
0.
x
vt
==
The graph is shown.
(b) Strategy
Use Eq. (44).
Solution
Find the distance the train traveled.
22
2
i
11
1
() (
0
)
()
2
1
(1.20 m s )(12.0 s)
86.4 m
2
xx
x
x
x
vt a t
t a t
a t
Δ = Δ+
Δ
=
Δ+
Δ
=
Δ
v
x
(m/s)
t
(s)
2
4
6
8
10
12
14
0
2
4
6
8 10 12
16
0
(c)
Strategy
Use Eq. (41).
Solution
Find the final speed of the train.
2
fi
f
f
0
, so
(1.20 m s )(12.0 s)
14.4 m s .
x
x
vvv
a
tv
a
t
−=−
=
Δ
=
Δ
=
=
17.
Strategy
Refer to the figure. Analyze graphically and algebraically.
Solution
Graphical analysis: Find the slope of the graph.
2
av,
40 m s
20 m s
5.0 m s
9.0 s
5.0 s
x
a
−
−
Algebraic solution:
2
a
v
,
a
v
,
40 m s
20 m s
, so
5.0 m s .
This is the end of the preview. Sign up
to
access the rest of the document.
 Fall '08
 ARIAS
 Acceleration

Click to edit the document details