perez (nap563) – HW01 – Tsoi – (58160)
1
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printout
should
have
24
questions.
Multiplechoice questions may continue on
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before answering.
001
10.0 points
A mass attached to a spring oscillates back
and forth as indicated in the position
vs.
time
plot below.
P
x
t
At point P, the mass has
1.
positive velocity and negative accelera
tion.
correct
2.
positive velocity and positive accelera
tion.
3.
negative velocity and negative accelera
tion.
4.
negative velocity and positive accelera
tion.
5.
zero velocity and zero acceleration.
6.
negative velocity and zero acceleration.
7.
zero velocity but is accelerating (posi
tively or negatively).
8.
positive velocity and zero acceleration.
Explanation:
The velocity is positive because the slope of
the curve at P is positive.
The acceleration
is negative because the curve is concave down
at P.
002
10.0 points
Consider three position curves between time
points
t
A
and
t
B
.
s
B
s
A
B
A
t
A
t
B
3
2
1
0
t
s
Choose
the
correct
relationship
among
quantities
v
1
,
v
2
, and
v
3
.
v
=
v
A
+
v
B
2
when
a
is a constant.
1.
v
1
>
v
2
>
v
3
2.
v
1
<
v
2
<
v
3
3.
v
1
=
v
2
=
v
3
correct
Explanation:
The average velocity of an object is
v
=
displacement
time
=
s
B

s
A
t
B

t
A
.
All three curves have exactly the same
change in position Δ
s
=
s
B

s
A
in exactly
the same time interval Δ
t
=
t
B

t
A
, so all
three average velocities are equal:
v
1
=
v
2
=
v
3
.
003
(part 1 of 2) 10.0 points
Consider the positiontime graph for a squir
rel running along a clothesline.
1
2
3
4
5
0
1
2
3
4

1

2
time (s)
position (m)
What is the squirrel’s displacement at the
time
t
= 4
.
0 s?
Your answer must be within
±
5.0%
Correct answer:

1 m.
Explanation:
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perez (nap563) – HW01 – Tsoi – (58160)
2
Let :
x
f
=

1 m
and
x
i
= 0 m
.
The displacement is
Δ
x
=
x
f

x
i
= (

1 m)

(0 m) =

1 m
.
004
(part 2 of 2) 10.0 points
What is the squirrel’s average velocity during
the time interval between 0.0 s and 4.0 s?
Your answer must be within
±
5.0%
Correct answer:

0
.
25 m
/
s.
Explanation:
Let :
Δ
t
= 4
.
0 s
.
v
avg
=
Δ
x
Δ
t
=

1 m
4 s
=

0
.
25 m
/
s
.
005
10.0 points
The graph shows the velocity
v
as a function
of time
t
for an object moving in a straight
line.
t
v
0
t
Q
t
R
t
S
t
P
Which graph shows the corresponding dis
placement
x
as a function of time
t
for the
same time interval?
1.
t
x
0
t
Q
t
R
t
S
t
P
2.
None of these graphs is correct.
3.
t
x
0
t
Q
t
R
t
S
t
P
4.
t
x
0
t
Q
t
R
t
S
t
P
5.
t
x
0
t
Q
t
R
t
S
t
P
6.
t
x
0
t
Q
t
R
t
S
t
P
7.
t
x
0
t
Q
t
R
t
S
t
P
correct
8.
t
x
0
t
Q
t
R
t
S
t
P
9.
t
x
0
t
Q
t
R
t
S
t
P
Explanation:
The displacement is the integral of the ve
locity with respect to time:
vectorx
=
integraldisplay
vectorv dt .
Because the velocity increases linearly from
zero at first, then remains constant, then de
creases linearly to zero, the displacement will
increase at first proportional to time squared,
then increase linearly, and then increase pro
portional to negative time squared.
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 Spring '08
 Turner
 Physics, Acceleration, Mass, Velocity, Correct Answer, m/s

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