Oslund (hno56) – Homework 1 – sai – (58015)
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001
(part 1 of 2) 10.0 points
While John is traveling along a straight inter
state highway, he notices that the mile marker
reads 241 km.
John travels until he reaches
the 148 km marker and then retraces his path
to the 172 km marker.
What is John’s resultant displacement from
the 241 km marker?
Correct answer:

69 km.
Explanation:
Let the initial position be
s
o
=
a
= 241 km,
the second position
b
= 148 km, and his final
position
s
f
=
c
= 172 km. Thus
Δ
s
=
s
f

s
o
= 172 km

241 km =

69 km
002
(part 2 of 2) 10.0 points
How far has he traveled?
Correct answer: 117 km.
Explanation:
The distance traveled is given by
d
=

b

a

+

c

b

=

148 km

241 km

+

172 km

148 km

= 117 km
003
(part 1 of 2) 10.0 points
Consider a moving object whose position
x
is plotted as a function of the time
t
on the
following figure:
2
4
6
2
4
6
t
x
I
II
III
Clearly, the object moved in di
ff
erent ways
during the time intervals denoted I, II and III
on the figure.
During these three intervals, when was the
object’s
speed
highest?
Caution:
Do not
confuse the speed with the velocity.
1.
During interval I.
2.
Same speed during each of the three in
tervals.
3.
During interval III.
correct
4.
During interval II.
5.
During intervals II and III (same speed
during those two intervals).
Explanation:
The velocity
v
is the slope of the
x
(
t
) curve;
the magnitude
v
=

v

of this slope is the
speed.
Looking at the picture we see that
the curve is steepest (in absolute magnitude)
during the interval III and that is when the
object had the highest speed.
004
(part 2 of 2) 10.0 points
During which interval(s) did the object’s ve
locity remain constant?
1.
During each of the three intervals.
cor
rect
2.
During interval I only.
3.
During none of the three intervals.
4.
During interval III only.
5.
During interval II only.
Explanation:
For each of the three intervals I, II or III,
the
x
(
t
) curve is linear, so its slope — the
velocity
v
— is constant during each interval.
Between
the intervals, the velocity did change
in a rather abrupt manner — but it did remain
constant
during
each interval.
005
(part 1 of 6) 5.0 points
Consider the following graph of motion.
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Oslund (hno56) – Homework 1 – sai – (58015)
2
0
1
2
3
4
5
0
10
20
30
40
50
Time (sec)
Distance (m)
How far did the object travel between 2 s
and 4 s?
1.
40 m
2.
50 m
3.
10 m
4.
30 m
5.
20 m
correct
Explanation:
The particle went from 40 m to 20 m, so
Δ
d
= 40 m

20 m =
20 m
.
006
(part 2 of 6) 5.0 points
The graph indicates
1.
decreasing velocity.
2.
constant velocity.
correct
3.
no motion.
4.
increasing velocity.
5.
constant position.
Explanation:
The slope of the graph is the same every
where, so the graph indicates constant posi
tive velocity.
007
(part 3 of 6) 5.0 points
What is the speed from 2 s to 4 s?
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 Spring '08
 Kaplunovsky
 Physics, Acceleration, Work, Velocity, Correct Answer, m/s, Miles per hour

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