1
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001 (part 1 of 2) 10.0 points
You’re driving on a straight road (in the +
x
direction) at a constant speed of 30 m
/
s. In
8 s, you speed up to 44 m
/
s to pass a truck.
Assuming that your car speeds up at a con
stant rate (constant force by the road on
the tires), what is your average
x
component
of velocity
v
avg
,X
during this maneuver?
Correct answer:
37 m
/
s.
Explanation:
Since the car speeds up at a constant
rate,
we
just
need
to
consider
the
endpoints, when the car was traveling
v
I
,X
= 30 m
/
s
and
v
F
,X
= 44 m
/
s
.
We calculate the average
x
component
of velocity as follows:
v
avg
,X
=
v
I
,X
+
v
F
,X
2
=
30 m
/
s + 44 m
/
s
2
=
37 m
/
s
.
3
(part 2 of 2) 10.0 points
How far do you go during this maneuver?
Correct answer:
296 m.
Explanation:
Again, the fact that the acceleration is con
stant simplifies things. We can just treat this
situation as though we were actually moving at
the average speed for the 8 s interval:
x
= =
v
avg
,X
t
=
(37 m
/
s)(8 s)
=
296 m
.
003 (part 1 of 3) 0.0 points
On a straight road (taken to be in the
+
x
direction) you drive for an hour at
v
1
,X
= 55 km
/
h
,
then quickly speed up to
v
2
,X
= 115 km
/
h
and drive for an additional two hours.
How far do you go (Δ
x
)?
Correct answer:
285 km.
Explanation:
To find the total distance traveled, we
have to consider the two speeds separately:
x
=
v
1
,X
t
1
+
v
2
,X
t
2
=
(55 km
/
h)(1 h) + (115 km
/
h)(2 h)
=
285 km
.
4
(part 2 of 3) 0.0 points
What is your average
x
component of
velocity
(
v
avg
,X
)?
Correct answer:
95 km
/
h.
Explanation:
We can’t simply take the arithmetic mean in
this case, because the car did not accelerate
at a constant rate between the initial and final
speeds. Instead, we have to take a weighted
average, considering the two speeds and how
long the car spent at each speed:
v
avg
,X
=
v
1
,X
t
1
+
v
2
,X
t
2
t
1
+
t
2
=
(55 km
/
h)(1 h) + (115 km
/
h)(2 h)
2
h + 1 h
=
95 km
/
h
.
5
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 Spring '08
 Turner
 Acceleration, Velocity, Ch2HW2 – florin

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