Version 001 – Summer Review #03 – tubman – (IBII20142015)
This print-out should have 35 questions.
Multiple-choice questions may continue on
the next column or page – find all choices
before answering.
Concept 20 P03
001 10.0 points
An oceanic depth-sounding vessel surveys the
ocean bottom with ultrasonic waves that
travel 1530 m/s in seawater.
How deep is the water directly below the
vessel if the time delay of the echo to the
ocean floor and back is 4 s?
Correct answer: 3060 m.
Explanation:
Let :
v = 1530 m/s and
∆t = 4 s.
The sound takes 2 s to reach the ocean floor
(and 2 s to return), so
d = v t = (1530 m/s) (2 s) = 3060 m .
Constant Velocity 02
002 10.0 points
An object is moving with constant velocity ~v .
Which situation is impossible in such a circumstance?
1. Four forces act on the object.
2. One force acts on the object. correct
3. Three forces act on the object.
1
Displacement Curve 02
003 (part 1 of 2) 10.0 points
Consider a moving object whose position x
is plotted as a function of the time t. The
object moved in different ways during the
time intervals denoted I, II and III on the
figure.
6
x
4
2
I
II
2
III
4
t
6
During these three intervals, when was the
object’s speed highest? Do not confuse the
speed with the velocity.
1. Same speed during each of the three intervals.
2. During interval I
3. During interval II
4. Same speed during intervals II and III
5. During interval III correct
Explanation:
The velocity v is the slope of the x(t) curve;
the magnitude v = |v| of this slope is the
speed. The curve is steepest (in absolute
magnitude) during the interval III and that is
when the object had the highest speed.
4. Two forces act on the object.
5. No forces act on the object.
Explanation:
Two or more vectors can sum to zero, but a
single nonzero vector cannot sum to zero, and
Newton’s Second Law requires that
X
~ = m~a = 0
F
in this case.
004 (part 2 of 2) 10.0 points
During which interval(s) did the object’s velocity remain constant?
1. During none of the three intervals
2. During interval II only
3. During each of the three intervals correct
Version 001 – Summer Review #03 – tubman – (IBII20142015)
4. During interval III only
5. During interval I only
Explanation:
For each of the three intervals I, II or III, the
x(t) curve is linear, so its slope (the velocity
v) is constant. Between the intervals, the
velocity changed in an abrupt manner, but it
did remain constant during each interval.
Average Speed on a Trip
005 (part 1 of 2) 10.0 points
A person travels by car from one city to another. She drives for 20.9 min at 78 km/h,
8.05 min at 114 km/h, 39.9 min at 48.3 km/h,
and spends 17.8 min along the way eating
lunch and buying gas.
Determine the distance between the cities
along this route.
2
Let : tother = 17.8 min .
The total time is
vav
t = t1 + t2 + t3 + tother
= 20.9 min + 8.05 min
+ 39.9 min + 17.8 min
= 1.44417 h , so
74.5845 km
x
= 51.6454 km/h .
= =
t
1.44417 h
Comparison of Average Velocity
007 10.0 points
The position-versus-time graph below describes the motion of three different bodies
(labeled 1, 2, 3).
x
xA
A
Correct answer: 74.5845 km.
1
2
3
Explanation:
xB
Let :
t1
v1
t2
v2
t3
v3
= 20.9 min ,
= 78 km/h ,
= 8.05 min ,
= 114 km/h ,
= 39.9 min , and
= 48.3 km/h .
x = x1 + x2 + x3
= v1 t1 + v2 t2 + v3 t3
= (78 km/h)(20.9 min)
+ (114 km/h)(8.05 min)
+ (48.3 km/h)(39.9 min)
= 74.5845 km .
006 (part 2 of 2) 10.0 points
Determine the average speed for the trip.
Correct answer: 51.6454 km/h.
Explanation:
B
tA
tB
t
Consider the average velocities of the three
bodies. Which of the following statements is
correct?
1. v̄1 > v̄2 > v̄3
2. v̄1 = v̄2 = v̄3 correct
3. v̄1 < v̄2 < v̄3
4. v̄1 > v̄2 and v̄3 > v̄2
Explanation:
Average velocity is
displacement
x − xA
v̄ =
= B
.
time
tB − tA
All three bodies have exactly same displacement in exactly same time, so the average
velocities are exactly equal: v̄1 = v̄2 = v̄3 .
Version 001 – Summer Review #03 – tubman – (IBII20142015)
Accelerating Object
008 10.0 points
If the acceleration of an object is zero at some
instant in time, what can be said about its
velocity at that time?
3
a
t
3.
correct
1. It is positive.
a
2. It is not changing at that time. correct
t
4.
3. Unable to determine.
4. It is negative.
a
5. It is zero.
a=
t
5.
Explanation:
The acceleration
∆v
=0
∆t
∆v = 0 .
a
t
6.
Acceleration Time Graph 01
009 (part 1 of 5) 10.0 points
a
Consider a toy car which can move to the
right (positive direction) or left on a horizontal surface along a straight line.
t
7.
car
a
v
8.
t
+
O
What is the acceleration-time graph if the
car moves toward the right (away from the
origin), speeding up at a steady rate?
a
1.
2. None of these graphs is correct.
t
Explanation:
Since the car speeds up at a steady rate,
the acceleration is a constant.
010 (part 2 of 5) 10.0 points
What is the acceleration-time graph if the car
moves toward the right, slowing down at a
steady rate?
Version 001 – Summer Review #03 – tubman – (IBII20142015)
a
011 (part 3 of 5) 10.0 points
What is the acceleration-time graph if the car
moves towards the left (toward the origin) at
a constant velocity?
t
1.
4
a
a
t
1.
t
2.
a
a
t
2.
t
3.
a
a
t
4.
t
3.
correct
correct
4. None of these graphs is correct.
a
t
5.
a
t
5.
a
t
6.
a
t
6.
a
7.
t
a
7.
8. None of these graphs is correct.
Explanation:
Since the car slows down, the acceleration
is in the opposite direction.
t
Version 001 – Summer Review #03 – tubman – (IBII20142015)
a
a
t
8.
5
t
7.
Explanation:
Since the car moves at a constant velocity,
the acceleration is zero.
a
t
8.
012 (part 4 of 5) 10.0 points
What is the acceleration-time graph if the car
moves toward the left, speeding up at a steady
rate?
a
t
1.
Explanation:
The same reason as Part 1.
a
t
2.
013 (part 5 of 5) 10.0 points
What is the acceleration-time graph if the car
moves toward the right at a constant velocity?
a
t
1.
correct
a
t
3.
a
t
2.
a
t
4.
a
t
3.
5. None of these graphs is correct.
a
a
6.
t
4.
t
Version 001 – Summer Review #03 – tubman – (IBII20142015)
a
t
5.
correct
6
Acceleration vs Time 01
015 (part 1 of 4) 10.0 points
Consider the plot below describing the acceleration of a particle along a straight line with
an initial position of 25 m and an initial velocity of −8 m/s.
4
a
3
t
6.
2
7. None of these graphs is correct.
a
t
8.
Explanation:
The same reason as Part 3.
Acceleration of a Vehicle
014 10.0 points
A car traveling in a straight line has a velocity
of 2.34 m/s at some instant. After 5.46 s, its
velocity is 12.2 m/s.
What is its average acceleration in this time
interval?
acceleration (m/s2 )
1
0
−1
−2
−3
−4
−5
−6
−7
−8
0
1
2
3
4 5
time (s)
6
7
8
9
What is the velocity at 1 s?
Correct answer: −5 m/s.
Explanation:
Correct answer: 1.80586 m/s2 .
Explanation:
Let :
Let :
v1 = 2.34 m/s ,
t = 5.46 s , and
v2 = 12.2 m/s .
v0 = −8 m/s ,
a1 = 3 m/s2 , and
∆t = 1 s .
The change in velocity is the area a ∆t between the acceleration curve and the time
axis.
The average acceleration is
aav
vf − vi
∆v
=
=
∆t
∆t
12.2 m/s − 2.34 m/s
=
5.46 s
= 1.80586 m/s2 .
vf = v0 + a1 ∆t
= −8 m/s + (3 m/s2 ) (1 s) = −5 m/s .
016 (part 2 of 4) 10.0 points
What is the position at 1 s?
Version 001 – Summer Review #03 – tubman – (IBII20142015)
7
so the position after 7 s is
Correct answer: 18.5 m.
Explanation:
Let :
1
a2 (∆t2 )2
2
= 31 m + (10 m/s) (1 s)
1
+ (−7 m/s2 ) (1 s)2 = 37.5 m .
2
x2 = x1 + v1 ∆t2 +
x0 = 25 m .
1
a1 ∆t2
2
= 25 m + (−8 m/s) (1 s)
1
+ (3 m/s2 ) (1 s)2 = 18.5 m .
2
x1 = x0 + v0 ∆t +
017 (part 3 of 4) 10.0 points
What is the velocity at 7 s?
Correct answer: 3 m/s.
Explanation:
Describing Motion
019 (part 1 of 2) 10.0 points
A car initially at rest on a straight road
accelerates according to the acceleration vs
time plot.
a
t3
Let :
∆t1 = 6 s ,
a2 = −7 m/s2 ,
∆t2 = 1 s .
t1
t
t2
and
The calculation is done in two parts, each
with constant acceleration.
The velocity after 6 s is
v1 = v0 + a1 ∆t1
= −8 m/s + (3 m/s2 )(6 s) = 10 m/s ,
so the velocity after 7 s is
v2 = v1 + a2 ∆t2
= 10 m/s + (−7 m/s2 )(1 s) = 3 m/s .
018 (part 4 of 4) 10.0 points
What is the position at 7 s?
Correct answer: 37.5 m.
Explanation:
The position after 6 s is
1
a1 (∆t1 )2
2
= (25 m) + (−8 m/s) (6 s)
1
+ (3 m/s2 ) (6 s)2 = 31 m ,
2
x1 = x0 + v0 ∆t1 +
t4
What is the best description of the motion
of the car? Take forward to be the positive
direction.
1. The car starts at rest, accelerates to a low
speed, comes to a stop, accelerates backwards
and cruises in reverse.
2. The car goes forward and then goes backward, ending where it started.
3. The car starts at rest, accelerates to a high
speed, cruises for a short while, decelerates to
a lower speed, then cruises. correct
4. The car goes backward and then goes
forward.
5. The car goes forward and then goes backward, ending behind where it started.
6. The car starts at rest, goes up to a high
speed, stops moving, travels backward, and
stops.
Explanation:
Version 001 – Summer Review #03 – tubman – (IBII20142015)
Use the acceleration to determine the velocity behavior:
1) v0 = 0 (initially at rest)
2) 0 < t < t1 : a = 0 (remains at rest)
3) t1 < t < t2 : a > 0 (accelerates forward)
4) t2 < t < t3 : a = 0 (constant speed)
5) t3 < t < t4 : a < 0 (decelerates)
6) t4 < t : a = 0 (constant speed)
v
t3
7.
t1
020 (part 2 of 2) 10.0 points
Which of the following graphs describes the
velocity vs time of the car?
t3
1.
t1
t4
t
t4
t
t2
v
t3
8.
v
8
t1
t2
t
t4
v
9.
t
t1
t2
t2
t3
t4
correct
2. None of these graphs is correct.
v
t2
3.
t3
t
t4
t1
v
t3
4.
t1
t4
t
t2
v
t2
5.
t3
t1
t
t4
Hewitt CP9 03 E07
021 10.0 points
Can an object reverse its direction of travel
while maintaining a constant acceleration?
1. No; if the acceleration is constant, the
direction of the speed remains unchanged.
2. Yes; a ball tossed upward reverses its direction of travel at its highest point. correct
v
6.
Explanation:
1) 0 < t < t1 : v0 = 0
2) t1 < t < t2 : velocity increases uniformly
3) t2 < t < t3 : constant velocity
4) t3 < t < t4 : velocity decreases uniformly
5) t4 < t : constant velocity
The velocity vs time graph can be found
by plotting the area under the acceleration vs
time curve.
t1
t
t2
t3
t4
3. No; the direction of the speed is always
the same as the direction of the acceleration.
4. Yes; a ball thrown toward a wall bounces
back from the wall.
5. All are wrong.
Explanation:
Velocity and acceleration need not be in the
Version 001 – Summer Review #03 – tubman – (IBII20142015)
same direction. When a ball is tossed upward
it experiences a constant acceleration directed
downward.
Hewitt CP9 03 E13
022 10.0 points
Which of the following is an example of something that undergoes acceleration while moving at constant speed?
1. None of these. An object that undergoes
an acceleration has to change its speed
2. A car moving straight backwards on the
road
Explanation:
At time t = 0, the velocity is 0.
024 (part 2 of 5) 10.0 points
What is the final velocity?
Correct answer: 25 m/s.
Explanation:
The final velocity is read from the graph
and is 5 times the vertical scale factor (5 m/s).
vf = 5 (5 m/s) = 25 m/s .
3. A football flying in the air
025 (part 3 of 5) 10.0 points
What is the final position?
4. A man standing in an elevator
Correct answer: 576.5 m.
5. A car making a circle in a parking lot
correct
Explanation:
When an object moves in a circular path
at constant speed its direction changes, so
its velocity changes, meaning it experiences
acceleration.
velocity (× 5 m/s)
Velocity vs Time 08
023 (part 1 of 5) 10.0 points
The scale on the horizontal axis is 5 s and on
the vertical axis 5 m/s. The initial position is
14 m.
9
Explanation:
The change in position is the area under
the velocity vs time graph, so
1
(9 t) (5 v)
2
1
= po + 9 (5 s) 5 (5 m/s)
2
= 576.5 m .
pf = po +
026 (part 4 of 5) 10.0 points
What acceleration is represented by the
graph?
6
Correct answer: 0.555556 m/s2 .
5
Explanation:
The acceleration is the slope of the velocity vs time graph, so
4
3
a=
2
1
0
∆v
5v − 0
5 (5 m/s)
=
=
∆t
9t− 0
9 (5 s)
= 0.555556 m/s2 .
0
1
2
3 4 5 6
time (× 5 s)
What is the initial velocity?
Correct answer: 0.
7
8
9
027 (part 5 of 5) 10.0 points
In which direction is the motion?
1. Unable to determine.
Version 001 – Summer Review #03 – tubman – (IBII20142015)
2. forward correct
3. negative
negative
3. backward
4. positive
positive
5. positive
negative
Explanation:
The velocity is positive, so the motion is
directed forward.
AP B 1993 MC 5
028 10.0 points
An object is released from rest on a planet
that has no atmosphere. The object falls
freely for 3 m in the first second.
What is the magnitude of the acceleration
due to gravity on the planet?
1. 3.0 m/s2
Explanation:
At the maximum elevation, the vertical velocity is zero. The acceleration is due to
gravity, which always acts down.
Acceleration of Falling Object
030 10.0 points
If you drop an object, it accelerates downward
at 9.8 m/s2 (in the absence of air resistance).
If, instead, you throw it downward, its
downward acceleration after release is
2. 12.0 m/s2
1. 9.8 m/s2 . correct
3. 6.0 m/s2 correct
2. less than 9.8 m/s2 .
4. 10.0 m/s2
3. more than 9.8 m/s2 .
Explanation:
The acceleration of all freely falling objects
is g = 9.8 m/s2 .
Let :
s = 3 m.
1 2
at
2
2s
2 (3 m)
a= 2 =
= 6 m/s2 .
t
(1.0 s)2
s=
AP M 1993 MC 19
029 10.0 points
An object is shot vertically upward into the
air with a positive initial velocity.
What correctly describes the velocity and
acceleration of the object at its maximum
elevation?
1. zero
2. zero
Ball M 01
031 10.0 points
A ball is thrown upward with an initial vertical speed of v0 to a maximum height of hmax .
bbb
bb
b b
b b
b
b
b
b
b
b
b
b
b
b
hmax
Explanation:
v0
5. 1.5 m/s2
Velocity
10
b
Acceleration
zero
negative correct
What is its maximum height hmax ? The
acceleration of gravity is g. Neglect air resis-
Version 001 – Summer Review #03 – tubman – (IBII20142015)
11
Explanation:
tance.
1. hmax =
2. hmax =
3. hmax =
4. hmax =
5. hmax =
6. hmax =
7. hmax =
8. hmax =
9. hmax =
v02
4g
3 v02
4g
√ 2
5v
√ 0
2 2g
v02
√
2g
2
v0
g
√ 2
3 v0
2g
5 v02
8g
√ 2
3v
√ 0
2 2g
v02
correct
2g
Let :
The distance an object falls from rest under
an acceleration a is
1 2
at
2
s
r
2h
2 (19 km)
=
= 110.716 s .
t=
a
(3.1 m/s2 )
h=
033 (part 2 of 2) 10.0 points
Find the magnitude of the velocity with which
the ball hits the crater floor.
Correct answer: 343.22 m/s.
Explanation:
Since the object falls from rest,
Explanation:
With up positive, a = −g and for the upward motion tf = 0, so for constant acceleration,
vf2 = v02 + 2 a yf − y0
0 = v02 − 2 g (hmax − 0)
v2
hmax = 0 .
2g
Ball Drop on Olympus
032 (part 1 of 2) 10.0 points
The tallest volcano in the solar system is the
19 km tall Martian volcano, Olympus Mons.
An astronaut drops a ball off the rim of the
crater and that the free fall acceleration of the
ball remains constant throughout the ball’s
19 km fall at a value of 3.1 m/s2 . (We assume
that the crater is as deep as the volcano is tall,
which is not usually the case in nature.)
Find the time for the ball to reach the crater
floor.
Correct answer: 110.716 s.
h = 19 km and
a = 3.1 m/s2 .
v2 = 2 a h
√
= q2 a h
v=
2 (3.1 m/s2 ) (19 km)
= 343.22 m/s .
Holt SF 02F 03
034 (part 1 of 2) 10.0 points
A tennis ball is thrown vertically upward with
an initial velocity of +7.0 m/s.
What will the ball’s velocity be when it
returns to its starting point? The acceleration
of gravity is 9.81 m/s2 .
Correct answer: −7 m/s.
Explanation:
Let :
vi = 7.0 m/s ,
a = −9.81 m/s2 ,
∆y = 0 m .
vf2 = vi2 + 2 a ∆y = vi2
vf = ±vi = ±7 m/s .
and
Version 001 – Summer Review #03 – tubman – (IBII20142015)
Since the direction is down, the final velocity
is −7 m/s .
035 (part 2 of 2) 10.0 points
How long will the ball take to reach its starting
point?
Correct answer: 1.42712 s.
Explanation:
vf = vi + a ∆t
vf − vi = a ∆t
vf − vi
−7 m/s − (7 m/s)
∆t =
=
a
−9.81 m/s2
= 1.42712 s .
12