muir (cem2577) – HW2 – Niu – (58205)
1
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printout
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have
30
questions.
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before answering.
001
10.0 points
On a drive from your ranch to Austin, you
wish to average 44 mph.
The distance from
your ranch to Austin is 110 miles. However,
at 55 miles (half way), you find you have
averaged only 33 mph.
What average speed must you maintain in
the remaining distance in order to have an
overall average speed of 44 mph?
Correct answer: 66 mph.
Explanation:
Let
t
denote the total time,
d
the total
distance, and
v
the average velocity over the
total distance. Let
t
1
denote the time over the
first half,
d
1
the first half distance, and
v
1
the
average velocity of the first half. Let
t
2
denote
the time over the second half,
d
2
the second
half distance, and
v
2
the average velocity of
the second half. The change in velocity occurs
at
d
1
, therefore
d
2
=
d
−
d
1
.
(1)
Average speed is
v
≡
Δ
d
Δ
t
=
d
t
. Thus
t
1
=
d
1
v
1
(2)
t
=
d
v
.
(3)
The total time
t
=
t
1
+
t
2
, so
t
2
=
t
−
t
1
=
d
v
−
d
1
v
1
(4)
since
d
2
=
v
2
t
2
v
2
=
d
2
t
2
≈
66 mph
.
Notice that the formula for average velocity
when acceleration is constant:
v
=
v
1
+
v
2
2
yields:
v
2
= 2
v
−
v
1
= 2(44 mph)
−
33 mph = 55 mph
an incorrect answer!!!! Do you know why this
formula is incorrect? What is the acceleration
at the halfway point?
What is the acceler
ation over the rest of the distance?
Is the
acceleration constant?
For constant acceler
ation
v
=
v
0
+
at
.
The graph of velocity
vs
time is a straight line.
For this problem the
graph of velocity
vs
time is a step function.
002
10.0 points
In order to qualify for the finals in a racing
event, a race car must achieve an average
speed of 243 km
/
h on a track with a total
length of 1490 m.
If a particular car covers the first half of the
track at an average speed of 205 km
/
h, what
minimum average speed must it have in the
second half of the event in order to qualify?
Correct answer: 298
.
293 km
/
h.
Explanation:
Let
t
t
be the maximum time to complete
the trip.
t
t
=
total distance
needed average speed
=
d
v
av
The time spent to the first half,
t
1
is
t
1
=
half distance
v
av
1
=
d/
2
v
av
1
Thus, the maximum time that can be spent
second half of the trip is
t
2
=
t
t
−
t
1
and the required average speed on the second
half is
v
av
2
=
d/
2
t
2
003
10.0 points
An electron travels 2
.
15 m in 5
.
57
×
10
−
8
s.
How fast is it traveling?
Correct answer: 3
.
85996
×
10
7
m
/
s.
Explanation:
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muir (cem2577) – HW2 – Niu – (58205)
2
Speed and velocity have units of distance
per unit time, so
v
=
s
t
=
2
.
15 m
5
.
57
×
10
−
8
s
= 3
.
85996
×
10
7
m
/
s
004
(part 1 of 2) 10.0 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
.
05 s to
t
2
= 4
.
69 s.
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 Spring '09
 NIU
 Acceleration, Velocity, Correct Answer, Muir

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