Answer, Key – Homework 2 – David McIntyre – 45123 – Mar 25, 2004
1
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Chapter 2 problems.
001
(part 1 oF 1) 0 points
The graph below shows the velocity
v
as a
Function oF time
t
For an object moving in a
straight line.
t
O
v
Which oF the Following graphs shows the
corresponding displacement
x
as a Function oF
time
t
For the same time interval?
1.
t
O
x
2.
t
O
x
3.
t
O
x
4.
t
O
x
correct
5.
t
O
x
Explanation:
The displacement is the integral oF the ve
locity with respect to time
~x
=
Z
~v dt .
Because the velocity increases linearly From
zero at frst, then remains constant, then de
creases linearly to zero, the displacement will
increase at frst proportional to time squared,
then increase linearly, and then increase pro
portional to negative time squared.
±rom these Facts, we can obtain the correct
answer.
002
(part 1 oF 2) 5 points
A particle moves according to the equation
x
= (10 m
/
s
2
)
t
2
where
x
is in meters and
t
is
in seconds.
±ind the average velocity For the time in
terval From
t
1
= 2
.
1 s to
t
2
= 3
.
95 s.
Correct answer: 60
.
5 m
/
s.
Explanation:
x
= (10 m
/
s
2
)
t
2
v
ave
=
Δ
x
Δ
t
=
(10 m
/
s
2
) (
t
2
2

t
2
1
)
t
2

t
1
=
(10 m
/
s
2
) [(3
.
95 s)
2

(2
.
1 s)
2
]
(3
.
95 s)

(2
.
1 s)
= 60
.
5 m
/
s
003
(part 2 oF 2) 5 points
±ind the average velocity For the time interval
From
t
1
= 2
.
1 s to
t
3
= 2
.
2 s.
Correct answer: 43 m
/
s.
Explanation:
v
ave
=
Δ
x
Δ
t
=
(10 m
/
s
2
) (
t
2
3

t
2
1
)
t
3

t
1
=
(10 m
/
s
2
) [(2
.
2 s)
2

(2
.
1 s)
2
]
(2
.
2 s)

(2
.
1 s)
= 43 m
/
s
004
(part 1 oF 1) 0 points
The acceleration oF an object, starting From
rest, is shown on the graph below.
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View Full DocumentAnswer, Key – Homework 2 – David McIntyre – 45123 – Mar 25, 2004
2
t(s)
a(m/s
2
)
5
5
1
2
3
4
5
Other than at t = 0, when is the velocity of
the object equal to zero?
1.
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 Spring '08
 RIJSSENBEEK
 Acceleration, Work, Velocity, Correct Answer, David McIntyre

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