hernandez (ejh742) – oldhomework 04 – Turner – (58120)
1
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printout
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have
12
questions.
Multiplechoice questions may continue on
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before answering.
001
(part 1 of 2) 10.0 points
The velocity
v
(
t
) of some particle is plotted
as a function of time on the graph below. The
scale on the horizontal axis is 1 s per division
and on the vertical axis 3 m
/
s per division.
Initially, the particle is at
x
0
= 17 m.
0
1
2
3
4
5
6
0
1
2
3
4
5
6
7
8
9
v
(
t
)
time (
×
1 s)
velocity (
×
3 m
/
s)
What is the position
x
of the particle at
time
t
= 4 s?
Correct answer: 47 m.
Explanation:
Looking at the
v
(
t
) plot we see that over
time
t
= 4 (1 s) = 4 s, the particle’s velocity
decreases from the initial
v
0
= 4 (3 m
/
s) =
12 m
/
s to final
v
f
= 1 (3 m
/
s) = 3 m
/
s
.
The
v
(
t
) line is straight, which indicates constant
deceleration rate, so the average velocity is
given by
¯
v
=
v
0
+
v
f
2
=
12 m
/
s + 3 m
/
s
2
= 7
.
5 m
/
s
,
the particle’s displacement is
Δ
x
=
t
·
¯
v
= (4 s) (7
.
5 m
/
s) = 30 m
,
and its final position
x
=
x
0
+ Δ
x
= 17 m + 30 m =
47 m
.
002
(part 2 of 2) 10.0 points
What is the particle’s acceleration?
Correct answer:

2
.
25 m
/
s
2
.
Explanation:
The average acceleration of the particle is
¯
a
=
Δ
v
Δ
t
=
v
f

v
0
t
=
3 m
/
s

12 m
/
s
4 s
=

2
.
25 m
/
s
2
Since the
v
(
t
) line is straight, the acceleration
is constant, so
a
= ¯
a
=

2
.
25 m
/
s
2
.
003
(part 1 of 2) 10.0 points
A speeder passes a parked police car at 25.7
m/s.
Instantaneously, the police car starts
from rest with a uniform acceleration of 2.45
m/s
2
.
a) How much time pases before the speeder
is overtaken by the police car?
Correct answer: 20
.
9796 s.
Explanation:
Basic Concepts:
For the speeder,
Δ
x
s
=
v
s
Δ
t
For the policeman,
v
i
= 0 m/s, so the equa
tion simplifies to
Δ
x
p
=
1
2
a
p
(Δ
t
)
2
Let :
v
s
= 25
.
7 m
/
s
v
i,p
= 0 m
/
s
a
p
= 2
.
45 m
/
s
2
.
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 Spring '08
 Turner
 Physics, Acceleration, Work, Velocity, Correct Answer

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