Question 1, chap 2, sect 2.
part 1 of 1
10 points
The graph below 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 of the following graphs shows the
corresponding displacement
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.
t
x
0
t
Q
t
R
t
S
t
P
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
correct
7.
None of these graphs are 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.
From these facts, we can obtain the correct
answer.
t
x
0
t
Q
t
R
t
S
t
P
Question 2, chap 2, sect 2.
part 1 of 2
10 points
You drive a car 9 h at 46 km
/
h, then 9 h at
66 km
/
h.
What is your average velocity?
Correct answer: 56 km
/
h (tolerance
±
1 %).
Explanation:
Basic Concept
Average velocity is defined by
v
av
=
Δ
d
Δ
t
Solution
If you drive for
t
hours at a speed of
v
1
and
for the same time at a speed of
v
2
, then
Δ
d
=
d
1
+
d
2
=
v
1
t
+
v
2
t
in a time of
Δ
t
=
t
1
+
t
2
= 2
t
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so that the average velocity is
v
av
=
v
1
t
+
v
2
t
2
t
=
v
1
+
v
2
2
=
(46 km
/
h) + (66 km
/
h)
2
=
56 km
/
h
.
Question 3, chap 2, sect 2.
part 2 of 2
10 points
What is your average velocity if you drive
a distance of 504 km at a speed of 46 km
/
h,
then the same distance at a speed of 66 km
/
h?
Correct answer: 54
.
2143
km
/
h (tolerance
±
1 %).
Explanation:
If you drive the distance
d
at
v
1
and the
same distance at
v
2
, then
Δ
d
=
d
1
+
d
2
= 2
d
and
Δ
t
=
t
1
+
t
2
=
d
v
1
+
d
v
2
so that the average velocity will be
v
av
=
2
d
d
v
1
+
d
v
2
·
v
1
v
2
v
1
v
2
=
2
d v
1
v
2
d v
2
+
d v
1
=
2
v
1
v
2
v
1
+
v
2
=
2 (46 km
/
h) (66 km
/
h)
(46 km
/
h) + (66 km
/
h)
=
54
.
2143 km
/
h
.
Question 4, chap 2, sect 2.
part 1 of 2
10 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.
Find the average velocity for the time in
terval from
t
1
= 2
.
16 s to
t
2
= 4
.
63 s.
Correct answer: 67
.
9 m
/
s (tolerance
±
1 %).
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
) [(4
.
63 s)
2

(2
.
16 s)
2
]
(4
.
63 s)

(2
.
16 s)
= 67
.
9 m
/
s
Question 5, chap 2, sect 2.
part 2 of 2
10 points
Find the average velocity for the time in
terval from
t
1
= 2
.
16 s to
t
3
= 2
.
26 s.
Correct answer: 44
.
2 m
/
s (tolerance
±
1 %).
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
.
26 s)
2

(2
.
16 s)
2
]
(2
.
26 s)

(2
.
16 s)
= 44
.
2 m
/
s
Question 6, chap 2, sect 2.
part 1 of 1
10 points
A car travels along a straight stretch of
road. It proceeds for 12
.
5 mi at 51 mi
/
h, then
25
.
1 mi at 45 mi
/
h, and finally 39
.
6 mi at
35
.
4 mi
/
h.
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 Fall '08
 Kaplunovsky
 Acceleration, Velocity, m/s, Helen –

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