angles test 2
Multiple Choice
Identify the choice that best completes the statement or answers the question.
____
1. A man starts his car from rest and accelerates at 1 m/s2 for 2 seconds. He then continues at a constant velocity
for 10 seconds until he sees a tree blocking the road and applies brakes. The car, decelerating at 1 m/s2, finally
comes to rest. Which of the following graphs represents the motion correctly?
a.
c.
v (m/s)
30
28
26
24
22
20
18
16
14
12
10
8
6
4
2
5
b.
10
10
____
____
20
25
30
35
40
t (s)
10
8
8
6
6
4
4
2
2
10
15
20
25 t (s)
v (m/s)
5
d.
v (m/s)
5
____
15
38
36
34
32
30
28
26
24
22
20
18
16
14
12
10
8
6
4
2
10
15
20
25
30
35 t (s)
v (m/s)
5
10
15
20
25 t (s)
2. The path of a projectile through space is called its:
a. equilibrant
c. range
b. torque
d. trajectory
3. A soldier throws a grenade horizontally from the top of a cliff. Which of the following curves best describes the
path taken by the grenade?
a. Circle
c. Hyperbola
b. Ellipse
d. Parabola
4. A stone is thrown horizontally from the top of a 25.00-m cliff. The stone lands at a distance of 40.00 m from the
edge of the cliff. What is the initial horizontal velocity of the stone?
a. 2.260 m/s
c. 17.70 m/s
b. 15.60 m/s
d. 22.05 m/s
____
____
____
____
____
5. A ball is thrown horizontally at 10.0 m/s from the top of a hill 50.0 m high. How far from the base of the cliff
would the ball hit the ground?
a. 23.6 m
c. 31.9 m
b. 26.4 m
d. 45.0 m
6. A ball is thrown horizontally from a hill 29.0 m high at a velocity of 4.00 m/s. Find the distance between the
base of the hill and the point where the ball hits the ground.
a. 2.43 m
c. 10.06 m
b. 9.73 m
d. 3.28 m
7. A player kicks a football at an angle of 30.0° above the horizontal. The football has an initial velocity of 20.0
m/s. Find the horizontal component of the velocity and the maximum height attained by the football.
a. 10.0 m/s, 17.6 m
c. 25.1 m/s, 7.40 m
b. 17.3 m/s, 5.10 m
d. 30.3 m/s, 9.50 m
8. A missile launches at a velocity of 30.0 m/s at an angle of 30.0° to the normal. What is the maximum height the
missile attains?
a. 11.5 m
c. 34.4 m
b. 27.5 m
d. 45.9 m
9. A player kicks a football with an initial velocity of 3.00 m/s at an angle of 60.0° above the horizontal. What is
the horizontal distance traveled by the football?
a. 0.312 m
c. 0.673 m
b. 0.397 m
d. 0.795 m
Short Answer
10. The velocity-time graph of a car’s motion is given below. Plot the corresponding acceleration-time graph.
v (m/s)
10
5
5
10
15
t (s)
–5
–10
Problem
11. The velocity-time graph of the motion of a particle is shown below. Calculate the total displacement of the
particle from 0 to 29 seconds.
v (m/s)
34
32
30
28
26
24
22
20
18
16
14
12
10
8
6
4
2
2
4
6
8 10 12 14 16 18 20 22 24 26 28 30 32 34 36
t (s)
12. A ball is thrown vertically upward with a speed of 1.53 m/s from a point 4.21 m above the ground. Calculate the
time in which the ball will reach the ground.
13. A ball is thrown vertically upward with a speed of 1.86 m/s from a point 3.82 m above the ground. Calculate the
time in which the ball will reach the ground.
14. The graph below shows how the velocity of a rolling ball changes with time. Calculate the acceleration of the
ball.
v (m/s)
10
9
8
7
6
5
4
3
2
1
2
4
6
8
10
12
14
16
18
t (s)
15. A car accelerates from rest at 5 m/s2 for 5 seconds. It moves with a constant velocity for some time, and then
decelerates at 5 m/s2 to come to rest. The entire journey takes 25 seconds. Plot the velocity-time graph of the
motion.
16. The graph below represents the velocity-time variation of a car’s motion.
v (m/s)
15
10
5
5
10
15
20
25
30
t (s)
Use the graph to find:
a) The acceleration of the car between t = 0 s and t = 5 s.
b) The acceleration of the car between t = 5 s and t = 10 s.
c) The acceleration of the car between t = 10 s and t = 20 s.
d) The acceleration of the car between t = 20 s and t = 30 s.
17. A hiker throws a ball at an angle of 21.0° above the horizontal from a hill 21.0 m high. The hiker’s height is
1.750 m. The magnitudes of the horizontal and vertical components of the velocity are 14.004 m/s and 5.376
m/s, respectively. Find the distance between the base of the hill and the point where the ball hits the ground.
(Consider the hiker’s height while calculating the answer.)(in other words the height is 21.0+1.750
18. Two grenades, A and B, are thrown horizontally with different speeds from the top of a cliff 70 m high. The
speed of A is 2.50 m/s and the speed of B is 3.40 m/s. Both grenades remain in air for 3.77 s. Assume that the
acceleration due to gravity is 9.86 m/s2. What is the distance between A and B if they are thrown along the same
straight line?
19. A bus travels from west to east with a velocity of 11.5 m/s. A marble rolls on the surface of the bus floor with a
velocity of 0.200 m/s north. What is the velocity of the marble relative to the road?
angles test 2
Answer Section
MULTIPLE CHOICE
1. ANS: B
Feedback
A
B
C
D
This graph does not show a period of constant velocity.
Correct!
The car starts from rest instead of at constant speed and the slopes of the different parts of
the graph are incorrect.
The car moves at a constant velocity for 12 seconds.
PTS: 1
DIF: 3
REF: Page 57
OBJ: 3.1.3 Create velocity-time graphs.
NAT: B.4
TOP: Create velocity-time graphs.
KEY: Velocity-time graph
MSC: 3
NOT: /A/ This graph does not show a period of constant velocity. /B/ Correct! /C/ The car starts from rest
instead of at constant speed and the slopes of the different parts of the graph are incorrect./D/ The car moves at
a constant velocity for 12 seconds.
2. ANS: D
A projectile is any object that has been given an initial thrust and moves through air. Its path through space is
called its trajectory.
Feedback
A
B
C
D
Equilibrant is a type of force.
Torque is the product of the force and length of the lever arm.
Range is the horizontal distance traveled by a projectile.
Correct!
PTS: 1
DIF: 1
REF: Page 147
OBJ: 6.1.1 Recognize that the vertical and horizontal motions of a projectile are independent.
NAT: B.4
TOP: Recognize that the vertical and horizontal motions of a projectile are independent.
KEY: Projectile, Trajectory
MSC: 1
NOT: /a/ Equilibrant is a type of force. /b/ Torque is the product of the force and length of the lever arm. /c/
Range is the horizontal distance traveled by a projectile. /d/ Correct!
3. ANS: D
Draw a motion diagram for the trajectory showing the downward acceleration. The velocity will have two
components, a horizontal and a vertical component. The combination of constant horizontal velocity and
uniform vertical acceleration produces a distinct trajectory.
Feedback
A
B
C
D
A circular path would mean that the grenade falls on the soldier!
The grenade cannot take an elliptical path as gravity would act on it.
The trajectory of the grenade cannot be a hyperbola as Earth would exert a gravitational
pull on it.
Correct!
PTS: 1
DIF: 1
REF: Page 148
OBJ: 6.1.3 Explain how the shape of the trajectory of a projectile depends upon the frame of reference from
which it is observed.
NAT: B.4
TOP: Explain how the shape of the trajectory of a projectile depends upon the frame of reference from which it
is observed.
KEY: Projectile, Trajectory
MSC: 1
NOT: /a/ A circular path would mean that the grenade falls on the soldier! /b/ The grenade cannot take an
elliptical path as gravity would act on it. /c/ The trajectory of the grenade cannot be a hyperbola as Earth would
exert a gravitational pull on it. /d/ Correct!
4. ANS: C
Solve the equation for the time the stone is in the air. Then solve the equation for the initial horizontal velocity
by substituting the values of time and the range.
Feedback
A
B
C
D
The time period is 2.260 s. Divide the range by the time period to get the initial horizontal
velocity.
Divide the range by the period, instead of multiplying the range by the period.
Correct!
You calculated the vertical velocity instead of the horizontal velocity.
PTS: 1
DIF: 3
REF: Page 149
OBJ: 6.1.2 Relate the height, time in the air, and initial vertical velocity of a projectile using its vertical
motion, then determine the range using the horizontal motion.;
NAT: B.4
TOP: Relate the height, time in the air, and initial vertical velocity of a projectile using its vertical motion, then
determine the range using the horizontal motion.; KEY:
Projectiles launched horizontally
MSC: 3
NOT: /a/ The time period is 2.260 s. Divide the range by the time period to get the initial horizontal velocity.
/b/ Divide the range by the period, instead of multiplying the range by the period. /c/ Correct! /d/ You calculated
the vertical velocity instead of the horizontal velocity.
5. ANS: C
Use the equation for the y-position, and solve the equation for the time the stone is in the air. Multiply the time
period with the horizontal velocity to obtain the horizontal distance (range).
Feedback
A
B
C
D
Use the equation for the y-position to find the time period.
Multiply the period with the horizontal velocity.
Correct!
The magnitude of the acceleration due to gravity is 9.80 m/s^2.
PTS: 1
DIF: 2
REF: Page 149
OBJ: 6.1.2 Relate the height, time in the air, and initial vertical velocity of a projectile using its vertical
motion, then determine the range using the horizontal motion.;
NAT: B.4
TOP: Relate the height, time in the air, and initial vertical velocity of a projectile using its vertical motion, then
determine the range using the horizontal motion.; KEY:
Projectiles launched horizontally
MSC: 3
NOT: /a/ Use the equation for the y-position to find the time period. /b/ Multiply the period with the horizontal
velocity. /c/ Correct! /d/ The magnitude of the acceleration due to gravity is 9.80 m/s^2.
6. ANS: B
Use the equation for the y-position, and solve the equation for the time the stone is in the air. Multiply the time
period with the horizontal velocity to obtain the distance.
Feedback
A
B
C
D
You calculated the time the stone is in the air. Multiply this value with the initial
horizontal velocity to find the range.
Correct!
Use the equation for the y-position to calculate how long the stone is in the air.
To calculate the time the stone is in the air, use the formula t^2 = –2 y/g.
PTS: 1
DIF: 2
REF: Page 149
OBJ: 6.1.2 Relate the height, time in the air, and initial vertical velocity of a projectile using its vertical
motion, then determine the range using the horizontal motion.;
NAT: B.4
TOP: Relate the height, time in the air, and initial vertical velocity of a projectile using its vertical motion, then
determine the range using the horizontal motion.; KEY:
Projectiles launched horizontally
MSC: 3
NOT: /a/ You calculated the time the stone is in the air. Multiply this value with the initial horizontal velocity
to find the range. /b/ Correct! /c/ Use the equation for the y-position to calculate how long the stone is in the air.
/d/ To calculate the time the stone is in the air, use the formula t^2 = –2 y/g.
7. ANS: B
Write expressions for the vertical and horizontal velocity components. Solve the velocity equation for the time
of maximum height. Substitute this time into the equation for vertical position to find the height.
Feedback
A
B
C
D
You calculated the vertical component instead of the horizontal component.
Correct!
The actual velocity is always more than its horizontal component.
Can the horizontal component of the velocity be more than the actual velocity?
PTS: 1
DIF: 3
REF: Page 150
OBJ: 6.1.2 Relate the height, time in the air, and initial vertical velocity of a projectile using its vertical
motion, then determine the range using the horizontal motion.;
NAT: B.4
TOP: Relate the height, time in the air, and initial vertical velocity of a projectile using its vertical motion, then
determine the range using the horizontal motion.; KEY:
Maximum range
MSC: 3
NOT: /a/You calculated the vertical component instead of the horizontal component. /b/ Correct! /c/ The actual
velocity is always more than its horizontal component. /d/ Can the horizontal component of the velocity be
more than the actual velocity?
8. ANS: C
Write the equations for the initial velocity components, the velocity components at time t, and the position in
both directions.
Feedback
A
B
C
D
Did you multiply the velocity with the cosine of the angle instead of the sine?
The time taken to reach maximum height can be calculated using the formula t = vyo/g,
where vyo is the vertical component of the velocity.
Correct!
Substitute the vertical component of the velocity in the equation for the maximum height.
PTS: 1
DIF: 2
REF: Page 150
OBJ: 6.1.2 Relate the height, time in the air, and initial vertical velocity of a projectile using its vertical
motion, then determine the range using the horizontal motion.;
NAT: B.4
TOP: Relate the height, time in the air, and initial vertical velocity of a projectile using its vertical motion, then
determine the range using the horizontal motion.; KEY:
Maximum height
MSC: 2
NOT: /a/ Did you multiply the velocity with the cosine of the angle instead of the sine? /b/ The time taken to
reach maximum height can be calculated using the formula t = vyo/g, where vyo is the vertical component of the
velocity. /c/ Correct! /d/ Substitute the vertical component of the velocity in the equation for the maximum
height.
9. ANS: D
Solve the vertical-position equation for the time at the end of the flight, when y = 0. Substitute that value of time
into the equation for horizontal distance to get the range.
Feedback
A
B
Solve the vertical-position equation for the time at the end of the flight when y = 0.
To find the range, multiply the horizontal component of velocity with the time obtained
when y = 0.
The vertical component of velocity is the product of the velocity and the sine of the angle
the football makes with the horizontal.
Correct!
C
D
PTS: 1
DIF: 2
REF: Page 150
OBJ: 6.1.2 Relate the height, time in the air, and initial vertical velocity of a projectile using its vertical
motion, then determine the range using the horizontal motion.;
NAT: B.4
TOP: Relate the height, time in the air, and initial vertical velocity of a projectile using its vertical motion, then
determine the range using the horizontal motion.; KEY:
Maximum range
MSC: 2
NOT: /a/ Solve the vertical-position equation for the time at the end of the flight when y = 0. /b/ To find the
range, multiply the horizontal component of velocity with the time obtained when y = 0. /c/ The vertical
component of velocity is the product of the velocity and the sine of the angle the football makes with the
horizontal. /d/ Correct!
SHORT ANSWER
10. ANS:
a (m/s^2)
6
5
4
3
2
1
–1
1
2
3
4
5
6
7
8
9
t (s)
–2
–3
–4
–5
–6
PTS: 1
DIF: 3
REF: Page 66
OBJ: 3.1.2 Relate velocity and acceleration to the motion of an object.
NAT: B.4
TOP: Relate velocity and acceleration to the motion of an object.
KEY: Relate velocity and acceleration
MSC: 3
PROBLEM
11. ANS:
532.5 m
PTS:
OBJ:
NAT:
KEY:
NOT:
12. ANS:
1.10 s
1
DIF: 3
REF: Page 65
3.2.1 Interpret position-time graphs for motion with constant acceleration.
B.4
TOP: Interpret position-time graphs for motion with constant acceleration.
Position-time graph
MSC: 3
The area under the graph gives the distance traveled.
PTS: 1
DIF: 3
REF: Page 72
OBJ: 3.3.2 Solve objects involving objects in free fall.
NAT: B.4
TOP: Solve objects involving objects in free fall.
KEY: Free fall
MSC: 3
NOT: The total time is the sum of the time taken by the ball to reach its topmost point and the time to come
down from the topmost point to the ground.
13. ANS:
1.093 s
PTS: 1
DIF: 3
REF: Page 72
OBJ: 3.3.2 Solve objects involving objects in free fall.
NAT: B.4
TOP: Solve objects involving objects in free fall.
KEY: Free fall
MSC: 3
NOT: The total time is the sum of the time taken by the ball to reach its topmost point and the time to come
down from the topmost point to the ground.
14. ANS:
4 m/s2
PTS: 1
NAT: B.4
MSC: 3
15. ANS:
DIF: 3
REF: Page 59
OBJ: 3.1.1 Define acceleration.
TOP: Define acceleration.
KEY: Acceleration
NOT: Acceleration is the rate of change of velocity.
v (m/s)
30
25
20
15
10
5
5
10
15
20
25
30
t (s)
PTS: 1
DIF: 3
REF: Page 59
OBJ: 3.1.3 Create velocity-time graphs.
NAT: B.4
TOP: Create velocity-time graphs.
KEY: Velocity-time graph
MSC: 3
NOT: The velocity-time graph is a straight line making an acute angle with the positive x-axis for accelerated
motion, a horizontal line for uniform motion, and a straight line making an obtuse angle with the positive x-axis
for decelerated motion.
16. ANS:
a) 0.80 m/s2
b) 0 m/s2
c) 1 m/s2
d) –1.6 m/s2
PTS: 1
DIF: 3
REF: Page 66
OBJ: 3.2.3 Apply graphical and mathematical relationships to solve problems related to constant acceleration.
NAT:
B.4
TOP: Apply graphical and mathematical relationships to solve problems related to constant acceleration.
KEY: Problems related to constant acceleration
MSC: 3
NOT: The acceleration can be obtained by calculating the slope of the velocity-time graph.
17. ANS:
38.8 m
PTS: 1
DIF: 3
REF: Page 150
OBJ: 6.1.2 Relate the height, time in the air, and initial vertical velocity of a projectile using its vertical
motion, then determine the range using the horizontal motion.;
NAT: B.4
TOP: Relate the height, time in the air, and initial vertical velocity of a projectile using its vertical motion, then
determine the range using the horizontal motion.; KEY:
Projectiles launched at an angle
MSC: 3
NOT: Solve the y-equation for y = 0 to find the time the ball is in the air. Multiply the time with the horizontal
component of the velocity to obtain the range.
18. ANS:
3.39 m.
PTS: 1
DIF: 3
REF: Page 149
OBJ: 6.1.2 Relate the height, time in the air, and initial vertical velocity of a projectile using its vertical
motion, then determine the range using the horizontal motion.;
NAT: B.4
TOP: Relate the height, time in the air, and initial vertical velocity of a projectile using its vertical motion, then
determine the range using the horizontal motion.; KEY:
Projectile, Trajectory
MSC: 3
NOT: Use the equation for the y-position to solve the equation for the time the objects are in the air.
19. ANS:
The marble is traveling 11.5 m/s at 1.00 north of east.
PTS:
OBJ:
TOP:
MSC:
1
DIF: 2
REF: Page 157
6.3.2 Solve relative-velocity problems.
NAT: B.4
Solve relative-velocity problems.
KEY: Relative velocity
3
NOT: Use the equation for the relative velocity of an object.