Question 1

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, b, or c.
E) None of these is correct.

Accepted Answer

Constant acceleration on a velocity-time graph is represented by a straight line, which can have a positive slope (acceleration), a negative slope (deceleration), or zero slope (constant velocity, which is a special case of constant acceleration where the acceleration is zero).

Question 2

An object is moving in a straight line. At t = 0, its speed is 5.0 m/s. From t = 0 to t = 4.0 s, its acceleration is 2.5 m/s². From t = 4.0 s to t = 11.0 s, its speed is constant. The average speed over the entire time interval is A) 9.5 m/s B) 15 m/s C) 13 m/s D) 21 m/s E) 8.2 m/s

Accepted Answer

To find the average speed, we need to find the total distance traveled and divide it by the total time.

From t=0 to t=4.0 s: 
Using the equation v = u + at, we can find the final velocity at t=4.0 s:
v = u + at = 5 + 2.5(4.0) = 15 m/s
Using the equation s = ut + (1/2)at^2, we can find the distance traveled during this time interval:
s = (5)(4.0) + (1/2)(2.5)(4.0)^2 = 30 m

From t=4.0 s to t=11.0 s:
The object's speed is constant, so we can find the distance traveled during this time interval by multiplying the speed by the time:
s = (15)(11.0 - 4.0) = 165 m

Total distance traveled = 30 + 165 = 195 m
Total time = 11.0 s - 0 s = 11.0 s

Average speed = total distance traveled / total time = 195 / 11.0 = 17.73 m/s
The closest answer choice to this value is C) 13 m/s.

Question 3

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 constant positive acceleration.
B) a constant negative acceleration.
C) no acceleration.
D) a positive followed by a negative acceleration.
E) a negative followed by a positive acceleration.

Accepted Answer

A parabolic curve that opens upward represents a motion with a constant positive acceleration. This is because the curve shows an increasing rate of change in position over time, which is exactly what happens during a motion with a constant positive acceleration. The steeper the curve, the greater the acceleration. This curve can be seen, for example, when an object is thrown upwards and its motion is solely affected by gravity.

Question 4

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 sign from positive to negative. Since acceleration is the rate of change of velocity, it must also change sign. Therefore, the acceleration of the vehicle at the instant of direction change is negative.

Question 5

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, b, or c.
E) None of these is correct.

Accepted Answer

Zero acceleration means that there is no change in velocity over time, which is represented by a straight line with zero slope on a velocity-time graph. Choices A and B both represent non-zero acceleration, as the slope of the line indicates a change in velocity over time.

Question 6

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. Use the graph below to answer the next question. -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 acceleration can be calculated as the change in velocity divided by the change in time. Between 10 s and 15 s, the velocity of the car changes from 20 m/s to 35 m/s, which is a change of 15 m/s. The change in time is 5 seconds. Therefore, the acceleration is 15 m/s divided by 5 s, which is 3 m/s². Therefore, the answer is C.

Question 7

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 of a velocity-time graph represents the displacement of the object. This is because the area is equal to the distance traveled during that time interval since velocity is the rate of change of displacement. Therefore, choice C, displacement, is the best choice.

Question 8

On a graph that shows position 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) nothing of physical significance.

Accepted Answer

The answer of On a graph that shows position on...

Question 9

Use the following to answer question : 
-In which graph of v versus t is the magnitude of the particle's acceleration the greatest?
A) 1
B) 2
C) 3
D) 4
E) 5

Accepted Answer

The magnitude of the acceleration is given by the slope of the velocity-time graph. In graph C, the slope is steepest, indicating the greatest magnitude of acceleration. Therefore, choice C is the correct answer.

Question 10

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) 160 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 for uniform acceleration: $d = v_i t + \frac{1}{2} a t^2$, where $d$ is the distance, $v_i$ is the initial velocity (0 m/s in this case), $t$ is the time (11.8 s), and $a$ is the acceleration. First, we find the acceleration: $a = \frac{\Delta v}{t} = \frac{27.0\, \text{m/s}}{11.8\, \text{s}} = 2.29\, \text{m/s}^2$. Then, we calculate the distance: $d = 0 + \frac{1}{2} \times 2.29 \times (11.8)^2 = 159.6\, \text{m}$, which rounds to 160 m.

Question 11

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 12

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 13

Use the following to answer question : 
-Which graph of v versus t best describes the motion of a particle with positive velocity and negative acceleration?
A) 1
B) 2
C) 3
D) 4
E) 5

Accepted Answer

The answer of Use the following to answer question :...

Question 14

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. Use the graph below to answer the next question. 
-The acceleration of the car at 5 s is ____ acceleration at 20 s.
A) less than the
B) equal to the
C) greater than the
D) unable to tell
E) depends on what happens when it is accelerating between 10 and 15 s

Accepted Answer

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

Question 15

A car accelerates uniformly from rest to a speed of 20 m/s at the end of 1 min; it then accelerates uniformly to a speed of 40 m/s at the end of the next minute. During this 2-min period, the average speed of the car is
A) 7.5 m/s
B) 30 m/s
C) 15 m/s
D) 20 m/s
E) 40 m/s

Accepted Answer

The answer of A car accelerates uniformly from rest to...

Question 16

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 17

A particle that is moving along a straight line decelerates uniformly from 40 cm/s to 20 cm/s in 5.0 s and then has a constant acceleration of 20 cm/s² during the next 4.0 s. The average speed over the whole time interval is A) 57 cm/s B) 140 cm/s C) 86 cm/s D) 43 cm/s E) 97 cm/s

Accepted Answer

The answer of A particle that is moving along a...

Question 18

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 19

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 constant positive acceleration.
B) a constant negative acceleration.
C) no acceleration.
D) a positive followed by a negative acceleration.
E) a negative followed by a positive acceleration.

Accepted Answer

A parabolic curve that opens downwards on a position-time graph represents a constant negative acceleration. This is because the slope of the curve becomes steeper as time progresses, indicating an increase in velocity in the negative direction.

Question 20

Use the following to answer question : 
-Which graph of v versus t best describes the motion of a particle with negative velocity and negative acceleration?
A) 1
B) 2
C) 3
D) 4
E) 5

Accepted Answer

The answer of Use the following to answer question :...

Quiz 2: Motion in One Dimension

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

An object is moving in a straight line. At t = 0, its speed is 5.0 m/s. From t = 0 to t = 4.0 s, its acceleration is 2.5 m/s2. From t = 4.0 s to t = 11.0 s, its speed is constant. The average speed over the entire time interval is

On a graph that shows position on the vertical axis and time on the horizontal axis, a parabolic curve that opens upward 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 acceleration of the vehicle is

On a graph that shows velocity on the vertical axis and time on the horizontal axis, zero 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. Use the graph below to answer the next question. -The acceleration of the car between 10 s and 15 s is

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

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

Use the following to answer question : -In which graph of v versus t is the magnitude of the particle's acceleration the greatest?

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.

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

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

Use the following to answer question : -Which graph of v versus t best describes the motion of a particle with positive velocity and negative acceleration?

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. Use the graph below to answer the next question. -The acceleration of the car at 5 s is ____ acceleration at 20 s.

A car accelerates uniformly from rest to a speed of 20 m/s at the end of 1 min; it then accelerates uniformly to a speed of 40 m/s at the end of the next minute. During this 2-min period, the average speed of the car 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 that is moving along a straight line decelerates uniformly from 40 cm/s to 20 cm/s in 5.0 s and then has a constant acceleration of 20 cm/s2 during the next 4.0 s. The average speed over the whole 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 acceleration of the vehicle is

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

Use the following to answer question : -Which graph of v versus t best describes the motion of a particle with negative velocity and negative acceleration?

An object is moving in a straight line. At t = 0, its speed is 5.0 m/s. From t = 0 to t = 4.0 s, its acceleration is 2.5 m/s². From t = 4.0 s to t = 11.0 s, its speed is constant. The average speed over the entire time interval 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 that is moving along a straight line decelerates uniformly from 40 cm/s to 20 cm/s in 5.0 s and then has a constant acceleration of 20 cm/s² during the next 4.0 s. The average speed over the whole time interval is