Adams, Russell – Homework 2 – Due: Sep 22 2006, midnight – Inst: Opyrchal, H
1
This
printout
should
have
15
questions.
Multiplechoice questions may continue on
the next column or page – find all choices
before answering.
The due time is Central
time.
001
(part 1 of 1) 10 points
A car travels along a straight stretch of road.
It proceeds for 17
.
8 mi at 58 mi
/
h, then 26 mi
at 43 mi
/
h, and finally 49
.
4 mi at 37
.
4 mi
/
h.
What is the car’s average velocity during
the entire trip?
Correct answer: 41
.
7487 mi
/
h.
Explanation:
The total distance the car traveled is
Δ
d
=
d
A
+
d
B
+
d
C
= 93
.
2 mi
The total time the car spent on the road is
Δ
t
=
d
A
v
A
+
d
B
v
B
+
d
C
v
C
=
17
.
8 mi
58 mi
/
h
+
26 mi
43 mi
/
h
+
49
.
4 mi
37
.
4 mi
/
h
= 2
.
2324 h
.
Hence the average velocity is
v
=
Δ
d
Δ
t
=
93
.
2 mi
2
.
2324 h
= 41
.
7487 mi
/
h
.
keywords:
002
(part 1 of 1) 10 points
A car, moving along a straight stretch of high
way, begins to accelerate at 0
.
0271 m
/
s
2
.
It
takes the car 49
.
4 s to cover 1 km.
How fast was the car going when it first
began to accelerate?
Correct answer: 19
.
5735 m
/
s.
Explanation:
We can describe this situation with the
equation
d
=
v
0
t
+
1
2
at
2
Given the time, distance and acceleration we
simply solve for the initial velocity
v
0
=
d
t

1
2
at
converting the distance from km to m to ob
tain the proper units.
keywords:
003
(part 1 of 8) 10 points
Consider the plot below describing motion
along a straight line with an initial position of
x
0
= 10 m.

2

1
0
1
2
3
4
5
6
7
0
1
2
3
4
5
6
7
8
9
time (s)
velocity (m/s)
What is the acceleration at 1 second?
Correct answer: 3
.
5 m
/
s
2
.
Explanation:
Basic Concepts:
The plot shows a curve
of the
velocity
versus
time
.
The acceleration is the slope of the velocity
curve
a
=
d v
dt
=
Δ
v
Δ
t
¶
.
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 Spring '08
 moro
 Physics, Work, Velocity, Correct Answer, m/s, 0 seconds, 19.5735 m/s, 25.9381 m/s

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