Question 1

A brass ball is shot vertically upward from the surface of an atmosphere-free planet with an initial speed of 20.0 m/s. One second later, the ball has an instantaneous velocity in the upward direction of 15.0 m/s. -What is the magnitude of the acceleration due to gravity on the surface of this planet? A)5.0 m/s² B)9.8 m/s² C)12 m/s² D)15 m/s² E)24 m/s²

Accepted Answer

The change in velocity is 5 m/s over 1 second, so the acceleration is 5 m/s².

Question 2

A brass ball is shot vertically upward from the surface of an atmosphere-free planet with an initial speed of 20.0 m/s. One second later, the ball has an instantaneous velocity in the upward direction of 15.0 m/s.
-How high does the ball rise?
A)70.0 m
B)10.0 m
C)50.0 m
D)20.0 m
E)40.0 m

Accepted Answer

The ball's velocity decreases by 5 m/s every second, indicating a constant acceleration (or deceleration) of -5 m/s². Using the formula v² = u² + 2as, where v = 0 m/s (at the highest point), u = 20 m/s, and a = -5 m/s², we find the maximum height s = 40 m.

Question 3

An object is moving along a straight line in the positive x direction. The graph shows its position from the starting point as a function of time. Various segments of the graph are identified by the letters A, B, C, and D.

-During which interval(s) did the object move in the negative x direction?
A)only during interval B
B)only during interval C
C)only during interval D
D)during both intervals C and D
E)The object never moved in the negative x direction.

Accepted Answer

The answer of An object is moving along a straight...

Question 4

A brass ball is shot vertically upward from the surface of an atmosphere-free planet with an initial speed of 20.0 m/s. One second later, the ball has an instantaneous velocity in the upward direction of 15.0 m/s.
-Determine the velocity of the ball when it returns to its original position. Note: assume the upward direction is positive.
A)+20 m/s
B)-20 m/s
C)+40 m/s
D)-40 m/s
E)zero m/s

Accepted Answer

First, let's use the equation for instantaneous velocity:

v = u + at

where v is final velocity, u is initial velocity, a is acceleration, and t is time. We can use this to find the acceleration of the brass ball one second after it is shot upwards:

15 m/s = 20 m/s + a(1 s)

a = -5 m/s^2

The negative sign means that the acceleration is in the downward direction.

Next, we can use the kinematic equation for displacement:

s = ut + (1/2)at^2

where s is displacement. We want to find the displacement of the ball when it returns to its original position, which means that s = 0. We can rearrange the equation to solve for the time it takes for the ball to reach its maximum height:

s = ut + (1/2)at^2
0 = 20 m/s(t) + (1/2)(-5 m/s^2)(t^2)
t = 4 s

Thus, it takes 4 seconds for the ball to reach its maximum height and return to its original position. We can use the equation for final velocity to find the velocity of the ball when it returns:

v = u + at
v = 20 m/s + (-5 m/s^2)(4 s)
v = -20 m/s

The negative sign means that the velocity of the ball when it returns is in the downward direction, so the best choice is B, -20 m/s.

Question 5

A automotive test driver travels due north in a prototype hybrid vehicle at 30 mi/h for 2 hours. She then reverses her direction and travels due south at 60 mi/h for 1 hour.
-What is the average velocity of the vehicle?
A)zero mi/h
B)40 mi/h, north
C)40 mi/h, south
D)45 mi/h, north
E)45 mi/h, south

Accepted Answer

The average velocity is the total displacement divided by the total time. The driver travels 60 miles north and then 60 miles south, resulting in zero displacement. Therefore, the average velocity is zero mi/h.

Question 6

A rock, dropped from rest near the surface of an atmosphere-free planet, attains a speed of 20.0 m/s after falling 8.0 meters.
-How long did it take the object to fall the 8.0 meters mentioned?
A)0.40 s
B)0.80 s
C)1.3 s
D)2.5 s
E)16 s

Accepted Answer

The time taken to fall a distance h is given by the formula t = sqrt(2h/g), where g is the acceleration due to gravity. Substituting the values given, we get t = sqrt(2*8/9.81) = 0.80 s. Therefore, the answer is B.

Question 7

A automotive test driver travels due north in a prototype hybrid vehicle at 30 mi/h for 2 hours. She then reverses her direction and travels due south at 60 mi/h for 1 hour.
-What is the average speed of the vehicle?
A)zero mi/h
B)30 mi/h
C)40 mi/h
D)50 mi/h
E)60 mi/h

Accepted Answer

To find the average speed of the vehicle, we need to use the formula:

Average speed = total distance / total time

The distance traveled north and south are equal since the driver returns to the starting point. Thus, we can find the total distance by doubling the distance traveled north, which is:

distance = speed × time
distance = 30 mi/h × 2 h
distance = 60 miles

The total time is the sum of the times traveled north and south:

total time = 2 h + 1 h
total time = 3 hours

Now we can plug in the values to find the average speed:

Average speed = total distance / total time
Average speed = 2 × 60 miles / 3 hours
Average speed = 40 mi/h

Therefore, the answer is choice C, 40 mi/h.

Question 8

A brass ball is shot vertically upward from the surface of an atmosphere-free planet with an initial speed of 20.0 m/s. One second later, the ball has an instantaneous velocity in the upward direction of 15.0 m/s.
-How long does it take the ball to reach its maximum height?
A)2.0 s
B)2.3 s
C)4.0 s
D)4.6 s
E)8.0 s

Accepted Answer

The answer of A brass ball is shot vertically upward...

Question 9

Starting from rest, a particle confined to move along a straight line is accelerated at a rate of 4 m/sS1U1P12S1S1P0.
-What is the speed of the particle after it has traveled 8 m?
A)4 m/s
B)8 m/s
C)30 m/s
D)60 m/s
E)100 m/s

Accepted Answer

Using the equation $v^2 = u^2 + 2as$, where $u = 0$, $a = 4 \, \text{m/s}^2$, and $s = 8 \, \text{m}$, we find $v = \sqrt{0 + 2 \times 4 \times 8} = 8 \, \text{m/s}$.

Question 10

In a race, Marcos runs 1.00 mile in 4.02 min, mounts a bicycle, and rides back to his starting point, which is also the finish line, in 3.02 min. What is the magnitude of Marcos' average velocity for the race?
A)zero mi/h
B)12.1 mi/h
C)14.9 mi/h
D)17.0 mi/h
E)19.9 mi/h

Accepted Answer

The average velocity is zero because the displacement for the entire race is zero; Marcos ends at the starting point.

Question 11

A small object is released from rest and falls 1.00 × 10S1U1P12S1S1P0 feet near the surface of the earth. Neglect air resistance. -How long will it take to fall through the 1.00 × 10²feet mentioned? A)2.49 s B)3.12 s C)4.50 s D)6.25 s E)10.0 s

Accepted Answer

Explanation:The distance fallen is given by:$$d = \frac{1}{2}gt^2$$where:d is the distance fallen (in feet)g is the acceleration due to gravity (32.2 ft/s^2)t is the time fallen (in seconds)Solving for t, we get:$$t = \sqrt{\frac{2d}{g}}$$Substituting the given values, we get:$$t = \sqrt{\frac{2(1.00 	imes 10^2 	ext{ ft})}{32.2 	ext{ ft/s}^2}} = 2.49 	ext{ s}$$Therefore, the correct answer is A.

Question 12

A brass ball is shot vertically upward from the surface of an atmosphere-free planet with an initial speed of 20.0 m/s. One second later, the ball has an instantaneous velocity in the upward direction of 15.0 m/s.
-How long is the ball in the air when it returns to its original position?
A)4.0 s
B)4.6 s
C)8.0 s
D)9.2 s
E)16 s

Accepted Answer

The answer of A brass ball is shot vertically upward...

Question 13

The figure shows the speed as a function of time for an object in free fall near the surface of the earth.  The object was dropped from rest in a long evacuated cylinder.
-Which one of the following statements best explains why the graph goes through the origin?
A)The object was in a vacuum.
B)The object was dropped from rest.
C)The velocity of the object was constant.
D)All v vs. t curves pass through the origin.
E)The acceleration of the object was constant.

Accepted Answer

When an object is dropped from rest, its initial velocity is zero. Therefore, the velocity-time graph starts at the origin.

Question 14

A rock, dropped from rest near the surface of an atmosphere-free planet, attains a speed of 20.0 m/s after falling 8.0 meters.
-How long would it take the object, falling from rest, to fall 16 m on this planet?
A)0.8 s
B)1.1 s
C)2.5 s
D)3.5 s
E)22 s

Accepted Answer

We can use the equation:

d = 1/2(gt^2)

where 
d = distance fallen = 16 m
g = acceleration due to gravity = 9.81 m/s^2 (assumed to be constant)
t = time elapsed

Rearranging:

t = √(2d/g)

t = √(2(16 m)/(9.81 m/s^2))

t ≈ 1.1 s

Therefore, the answer is choice B.

Question 15

Starting from rest, a particle confined to move along a straight line is accelerated at a rate of 4 m/sS1U1P12S1S1P0. -Which statement accurately describes the motion of the particle? A)The particle travels 4 meters during each second. B)The particle travels 4 meters during the first second only. C)The speed of the particle increases by 4 m/s during each second. D)The acceleration of the particle increases by 4 m/s² during each second. E)The final velocity of the particle will be proportional to the distance that the particle covers.

Accepted Answer

The statement "The speed of the particle increases by 4 m/s during each second" accurately describes the motion of the particle. This is because the particle is experiencing a constant acceleration of 4 m/s2, which means its speed increases by 4 m/s for every second of motion. Option A is incorrect because the distance the particle travels in each second will increase as its speed increases. Option B is incorrect because the particle's speed continues to increase after the first second. Option D is incorrect because the acceleration of the particle is constant, not increasing. Option E is incorrect because the final velocity of the particle depends on the time of motion and the acceleration, not the distance covered.

Question 16

A small object is released from rest and falls 1.00 × 10S1U1P12S1S1P0 feet near the surface of the earth. Neglect air resistance. -Approximately how fast will the object be moving after falling through the 1.00 × 10² feet mentioned? A)9.8 ft/s B)40 ft/s C)80 ft/s D)160 ft/s E)320 ft/s

Accepted Answer

We can use the equation for gravitational potential energy and kinetic energy to solve for the final velocity.

Gravitational potential energy at the top of the fall = mgh = m(32.2 ft/s²)(1.00 × 10¹² ft) = 3.22 × 10¹³m ft²/s²

At the bottom of the fall, all the potential energy has been converted to kinetic energy, so

Kinetic energy at bottom of the fall = 1/2 mv²

Equating the two:

1/2 mv² = 3.22 × 10¹³m ft²/s²

v² = 6.44 × 10¹³ ft²/s²

v = √(6.44 × 10¹³) ft/s

v ≈ 80 ft/s

Therefore, the best choice is C: 80 ft/s.

Question 17

The rate at which the acceleration of an object changes with time is called a jerk. What is the dimension of the jerk?
A) 
B) 
C) 
D) 
E)

Accepted Answer

The answer of The rate at which the acceleration of...

Question 18

Starting from rest, a particle confined to move along a straight line is accelerated at a rate of 4 m/sS1U1P12S1S1P0.
-After 10 seconds, how far will the particle have traveled?
A)20 m
B)40 m
C)100 m
D)200 m
E)400 m

Accepted Answer

The distance traveled by an object starting from rest with constant acceleration is given by the formula: $ s = \frac{1}{2} a t^2 $. Here, $ a = 4 \, \text{m/s}^2 $ and $ t = 10 \, \text{s} $. Plugging in these values, $ s = \frac{1}{2} \times 4 \times 10^2 = 200 \, \text{m} $.

Question 19

A rock, dropped from rest near the surface of an atmosphere-free planet, attains a speed of 20.0 m/s after falling 8.0 meters. -What is the magnitude of the acceleration due to gravity on the surface of this planet? A)0.40 m/s² B)1.3 m/s² C)2.5 m/s² D)25 m/s² E)160 m/s²

Accepted Answer

Using the kinematic equation $ v^2 = u^2 + 2as $, where $ v = 20.0 \, \text{m/s} $, $ u = 0 \, \text{m/s} $, and $ s = 8.0 \, \text{m} $, we solve for $ a $: $ 20^2 = 0 + 2a(8) $, giving $ a = 25 \, \text{m/s}^2 $.

Question 20

A rock, dropped from rest near the surface of an atmosphere-free planet, attains a speed of 20.0 m/s after falling 8.0 meters.
-Determine the speed of the object after falling from rest through 16 m on this planet.
A)28 m/s
B)32 m/s
C)56 m/s
D)64 m/s
E)320 m/s

Accepted Answer

The answer of A rock, dropped from rest near the...

Quiz 2: Kinematics in One Dimension

A brass ball is shot vertically upward from the surface of an atmosphere-free planet with an initial speed of 20.0 m/s. One second later, the ball has an instantaneous velocity in the upward direction of 15.0 m/s. -How high does the ball rise?

An object is moving along a straight line in the positive x direction. The graph shows its position from the starting point as a function of time. Various segments of the graph are identified by the letters A, B, C, and D. -During which interval(s) did the object move in the negative x direction?

A automotive test driver travels due north in a prototype hybrid vehicle at 30 mi/h for 2 hours. She then reverses her direction and travels due south at 60 mi/h for 1 hour. -What is the average velocity of the vehicle?

A rock, dropped from rest near the surface of an atmosphere-free planet, attains a speed of 20.0 m/s after falling 8.0 meters. -How long did it take the object to fall the 8.0 meters mentioned?

A automotive test driver travels due north in a prototype hybrid vehicle at 30 mi/h for 2 hours. She then reverses her direction and travels due south at 60 mi/h for 1 hour. -What is the average speed of the vehicle?

A brass ball is shot vertically upward from the surface of an atmosphere-free planet with an initial speed of 20.0 m/s. One second later, the ball has an instantaneous velocity in the upward direction of 15.0 m/s. -How long does it take the ball to reach its maximum height?

Starting from rest, a particle confined to move along a straight line is accelerated at a rate of 4 m/s². -What is the speed of the particle after it has traveled 8 m?

In a race, Marcos runs 1.00 mile in 4.02 min, mounts a bicycle, and rides back to his starting point, which is also the finish line, in 3.02 min. What is the magnitude of Marcos' average velocity for the race?

A small object is released from rest and falls 1.00 × 10² feet near the surface of the earth. Neglect air resistance. -How long will it take to fall through the 1.00 × 10²feet mentioned?

A brass ball is shot vertically upward from the surface of an atmosphere-free planet with an initial speed of 20.0 m/s. One second later, the ball has an instantaneous velocity in the upward direction of 15.0 m/s. -How long is the ball in the air when it returns to its original position?

The figure shows the speed as a function of time for an object in free fall near the surface of the earth. The object was dropped from rest in a long evacuated cylinder. -Which one of the following statements best explains why the graph goes through the origin?

A rock, dropped from rest near the surface of an atmosphere-free planet, attains a speed of 20.0 m/s after falling 8.0 meters. -How long would it take the object, falling from rest, to fall 16 m on this planet?

Starting from rest, a particle confined to move along a straight line is accelerated at a rate of 4 m/s². -Which statement accurately describes the motion of the particle?

A small object is released from rest and falls 1.00 × 10² feet near the surface of the earth. Neglect air resistance. -Approximately how fast will the object be moving after falling through the 1.00 × 10² feet mentioned?

The rate at which the acceleration of an object changes with time is called a jerk. What is the dimension of the jerk?

Starting from rest, a particle confined to move along a straight line is accelerated at a rate of 4 m/s². -After 10 seconds, how far will the particle have traveled?

A rock, dropped from rest near the surface of an atmosphere-free planet, attains a speed of 20.0 m/s after falling 8.0 meters. -What is the magnitude of the acceleration due to gravity on the surface of this planet?

A rock, dropped from rest near the surface of an atmosphere-free planet, attains a speed of 20.0 m/s after falling 8.0 meters. -Determine the speed of the object after falling from rest through 16 m on this planet.