Question 1

For uniformly accelerated motion, which of the following quantities must be zero?
A) the initial velocity
B) the initial displacement
C) the rate of change of the acceleration
D) the rate of change of the velocity
E) the rate of change of the displacement

Accepted Answer

For uniformly accelerated motion, the acceleration is constant, meaning the rate of change of the acceleration must be zero.

Question 2

On a graph that shows position on the vertical axis and time on the horizontal axis, a parabolic curve that opens upward represents
A) a motion with constant positive acceleration.
B) a motion with constant negative acceleration.
C) a motion with no acceleration.
D) a motion with positive acceleration followed by a motion with negative acceleration.
E) a motion with negative acceleration followed by a motion with positive acceleration.

Accepted Answer

A parabolic curve that opens upward represents a motion with constant positive acceleration. This is because the slope of the curve, which represents the velocity, increases steadily over time. The increasing slope corresponds to a constantly increasing velocity, and since acceleration is the rate of change of velocity, a constantly increasing velocity corresponds to a positive acceleration.

Question 3

On a graph that shows position on the vertical axis and time on the horizontal axis, a parabolic curve that opens downward represents
A) a motion with constant positive acceleration.
B) a motion with constant negative acceleration.
C) a motion with no acceleration.
D) a motion with a positive acceleration followed by a motion with negative acceleration.
E) a motion with negative acceleration followed by a motion with positive acceleration.

Accepted Answer

A downward-opening parabolic curve on a position-time graph represents a motion with a constant negative acceleration. As time increases, the position decreases at an increasing rate, causing the curve to become steeper. This is only possible if there is a negative acceleration acting on the object.

Question 4

A Triumph sports car starts at rest and accelerates uniformly to a speed of 27.0 m/s in 11.8 s. Calculate the distance the car travels during this time interval.
A) 159 m
B) 320 m
C) 1.90 km
D) 640 m
E) 350 m

Accepted Answer

The distance traveled by the car can be calculated using the formula:distance = (initial velocity + final velocity) * time / 2Since the car starts at rest, the initial velocity is 0 m/s. The final velocity is 27.0 m/s and the time is 11.8 s. Plugging these values into the formula, we get:distance = (0 m/s + 27.0 m/s) * 11.8 s / 2 = 159 mTherefore, the distance the car travels during this time interval is 159 m.

Question 5

On a graph that shows velocity on the vertical axis and time on the horizontal axis, the area under the curve represents
A) average acceleration.
B) average velocity.
C) displacement.
D) average speed.
E) no useful physical quantity.

Accepted Answer

The area under the curve on a velocity vs time graph represents the displacement of the object. This is because displacement is equal to the area under the curve on a velocity vs time graph. Therefore, the answer is C, displacement.

Question 6

A vehicle is traveling in the +x direction to x = 100 m. It then reverses direction. At the instant when it changes direction, the velocity of the vehicle is
A) positive.
B) negative.
C) zero.
D) positive then negative.
E) negative then positive.

Accepted Answer

At the instant a vehicle changes direction, its velocity is zero because it momentarily stops before reversing direction.

Question 7

A particle decelerates uniformly from a speed of 30 cm/s to rest in a time interval of 5.0 s. It then has a uniform acceleration of 10 cm/s² for another 5.0 s. The particle moves in the same direction along a straight line. The average speed over the whole time interval is A) 20 cm/s. B) 35 cm/s. C) 38 cm/s. D) 100 cm/s. E) 12 cm/s.

Accepted Answer

The answer of A particle decelerates uniformly from a speed...

Question 8

A particle accelerates uniformly from a speed of 30 cm/s to 40 cm/s in 5 s and thereafter moves at a constant speed of 40 cm/s for an additional 3 s. The average speed over this total time interval is
A) 35 cm/s.
B) 27 cm/s.
C) 0.45 cm/s.
D) 37 cm/s.
E) 73 cm/s.

Accepted Answer

The answer of A particle accelerates uniformly from a speed...

Question 9

A vehicle is traveling in the -x direction to x = 100 m. It then reverses direction. At the instant when it changes direction, the velocity n of the vehicle is
A) positive.
B) negative.
C) zero.
D) positive then negative.
E) negative then positive.

Accepted Answer

At the instant a vehicle changes direction, its velocity is zero because it momentarily stops before reversing direction.

Question 10

An object moves along the horizontal axis as shown on the diagram.  At which point or points is its acceleration zero?
A) A and E
B) B, D, and E
C) C only
D) E only
E) B and D

Accepted Answer

The answer of An object moves along the horizontal axis...

Question 11

A Lamborghini sports car can accelerate from zero to 60 mph in 4 seconds. It can decelerate from 60 mph to rest in 36.6 m. What is the ratio of the magnitudes of average acceleration over average deceleration? 1 mile = 1600 m) A) 1.74 * 10^-5 B) 1.47 C) 0.682 D) 0.0114 E) 0.688

Accepted Answer

The average acceleration is:$$a_a = \frac{60 mph}{4 s} = 15 mph/s$$Converting to m/s:$$a_a = 15 mph/s * (1600 m/1 mile) * (1 h/3600 s) = 6.71 m/s^2$$The average deceleration is:$$a_d = \frac{60 mph}{36.6 m} = 1.64 mph/s$$Converting to m/s:$$a_d = 1.64 mph/s * (1600 m/1 mile) * (1 h/3600 s) = 0.732 m/s^2$$The ratio of the magnitudes of average acceleration over average deceleration is:$$\frac{|a_a|}{|a_d|} = \frac{6.71 m/s^2}{0.732 m/s^2} = 9.17$$Therefore, the correct answer is C) 0.682.

Question 12

If we assume that a spaceship could accelerate from rest at a constant rate of 9.81 m/s², then how long would it take to reach 1% of the speed of light? Assume the speed of light = 3.0 *10⁸ m/s.) A) 1.8 days B) 3.5 days C) 8.6 h D) 1.5 days E) 7.1 days

Accepted Answer

We can use the equation: 
v = at
where v is the final velocity, a is the acceleration, and t is the time. 
We want to solve for t when v = 0.01c (1% of the speed of light), so we have: 
0.01c = (9.81 m/s^2)t 
Solving for t: 
t = 0.01c/9.81 m/s^2 
t = (0.01 x 3.0 x 10^8 m/s) / 9.81 m/s^2 
t ≈ 305,810 seconds 
Converting to days: 
t ≈ 305,810 seconds x (1 min/60 seconds) x (1 hour/60 min) x (1 day/24 hours) 
t ≈ 3.54 days 
Therefore, the answer is B) 3.5 days.

Question 13

A vehicle is traveling in the -x direction to x = 100 m. It then reverses direction. At the instant when it changes direction, the acceleration of the vehicle is
A) positive.
B) negative.
C) zero.
D) positive then negative.
E) negative then positive.

Accepted Answer

When the vehicle changes direction, its velocity changes from positive to negative. Therefore, the acceleration must be in the negative direction to cause this change in velocity. Thus, the correct answer is B, negative.

Question 14

A vehicle is traveling in the +x direction to x = 100 m. It then reverses direction. At the instant when it changes direction, the acceleration of the vehicle is
A) positive.
B) negative.
C) zero.
D) positive then negative.
E) negative then positive.

Accepted Answer

When the vehicle changes direction, it goes from moving in the positive x direction to moving in the negative x direction. This change in direction means that the velocity of the vehicle is changing, which indicates that there must be an acceleration acting on the object. Since the acceleration is in the negative direction (opposite to the direction of motion), the answer is B) negative.

Question 15

A car and a truck, starting from rest, have the same acceleration, but the truck accelerates for twice the length of time. Compared with the car, the truck will travel
A) twice as far.
B) three times as far.
C) 1.4 times as far.
D) four times as far.
E) one-half as far.

Accepted Answer

The answer of A car and a truck, starting from...

Question 16

The distance traveled by a car in the x-direction is shown. When the car changes speed for t = 10 s to 15 s, it does so uniformly.  The acceleration of the car at 5 s ____ acceleration at 20 s.
A) is less than the
B) is equal to the
C) is greater than the
D) cannot be determined for fore. depends on what happens when it is accelerating between 10 and 15 s

Accepted Answer

Since the car changes speed uniformly between t=10s and t=15s, the acceleration during that time period will also be constant. Therefore, the acceleration at t=5s and t=20s will be equal to the acceleration during the uniform speed change, which is the same at any point during that time period.

Question 17

On a graph that shows velocity on the vertical axis and time on the horizontal axis, zero acceleration is represented by
A) a straight line with a positive slope.
B) a straight line with a negative slope.
C) a straight line with zero slope.
D) either a positive, negative, or zero slope.
E) a curved line.

Accepted Answer

Zero acceleration means there is no change in velocity, which indicates a constant velocity. A straight line with a zero slope represents constant velocity, as the velocity is not changing with time.

Question 18

On a graph that shows velocity on the vertical axis and time on the horizontal axis, constant acceleration is represented by
A) a straight line with a positive slope.
B) a straight line with a negative slope.
C) a straight line with zero slope.
D) either a positive, negative, or zero slope.
E) None of these are correct.

Accepted Answer

A straight line with a positive slope represents constant acceleration because velocity is increasing at a constant rate over time.

Question 19

The distance traveled by a car in the x-direction is shown. When the car changes speed for t = 10 s to 15 s, it does so uniformly. The acceleration of the car between 10 s and 15 s is A) 1 m/s². B) 2 m/s². C) 3 m/s². D) 4 m/s². E) 5 m/s².

Accepted Answer

The answer of The distance traveled by a car in...

Question 20

On a graph that shows acceleration on the vertical axis and time on the horizontal axis, the area under the curve represents
A) average acceleration.
B) average velocity.
C) change in velocity.
D) displacement.
E) total distance.

Accepted Answer

The area under an acceleration vs. time graph represents the change in velocity, as acceleration is the rate of change of velocity with respect to time.

Quiz 2: Linear Motion

For uniformly accelerated motion, which of the following quantities must be zero?

On a graph that shows position on the vertical axis and time on the horizontal axis, a parabolic curve that opens upward represents

On a graph that shows position on the vertical axis and time on the horizontal axis, a parabolic curve that opens downward represents

A Triumph sports car starts at rest and accelerates uniformly to a speed of 27.0 m/s in 11.8 s. Calculate the distance the car travels during this time interval.

On a graph that shows velocity on the vertical axis and time on the horizontal axis, the area under the curve represents

A vehicle is traveling in the +x direction to x = 100 m. It then reverses direction. At the instant when it changes direction, the velocity of the vehicle is

A particle decelerates uniformly from a speed of 30 cm/s to rest in a time interval of 5.0 s. It then has a uniform acceleration of 10 cm/s2 for another 5.0 s. The particle moves in the same direction along a straight line. The average speed over the whole time interval is

A particle accelerates uniformly from a speed of 30 cm/s to 40 cm/s in 5 s and thereafter moves at a constant speed of 40 cm/s for an additional 3 s. The average speed over this total time interval is

A vehicle is traveling in the -x direction to x = 100 m. It then reverses direction. At the instant when it changes direction, the velocity n of the vehicle is

An object moves along the horizontal axis as shown on the diagram. At which point or points is its acceleration zero?

A Lamborghini sports car can accelerate from zero to 60 mph in 4 seconds. It can decelerate from 60 mph to rest in 36.6 m. What is the ratio of the magnitudes of average acceleration over average deceleration? 1 mile = 1600 m)

If we assume that a spaceship could accelerate from rest at a constant rate of 9.81 m/s2, then how long would it take to reach 1% of the speed of light? Assume the speed of light = 3.0 *108 m/s.)

A vehicle is traveling in the -x direction to x = 100 m. It then reverses direction. At the instant when it changes direction, the acceleration of the vehicle is

A vehicle is traveling in the +x direction to x = 100 m. It then reverses direction. At the instant when it changes direction, the acceleration of the vehicle is

A car and a truck, starting from rest, have the same acceleration, but the truck accelerates for twice the length of time. Compared with the car, the truck will travel

The distance traveled by a car in the x-direction is shown. When the car changes speed for t = 10 s to 15 s, it does so uniformly. The acceleration of the car at 5 s ____ acceleration at 20 s.

On a graph that shows velocity on the vertical axis and time on the horizontal axis, zero acceleration is represented by

On a graph that shows velocity on the vertical axis and time on the horizontal axis, constant acceleration is represented by

The distance traveled by a car in the x-direction is shown. When the car changes speed for t = 10 s to 15 s, it does so uniformly. The acceleration of the car between 10 s and 15 s is

On a graph that shows acceleration on the vertical axis and time on the horizontal axis, the area under the curve represents

A particle decelerates uniformly from a speed of 30 cm/s to rest in a time interval of 5.0 s. It then has a uniform acceleration of 10 cm/s² for another 5.0 s. The particle moves in the same direction along a straight line. The average speed over the whole time interval is

If we assume that a spaceship could accelerate from rest at a constant rate of 9.81 m/s², then how long would it take to reach 1% of the speed of light? Assume the speed of light = 3.0 *10⁸ m/s.)