Question 1

Two cars are traveling at the same speed and hit the brakes at the same time. Car A decelerates (decreases its velocity) at twice the rate of car B. If car B takes time T to stop, how long does it take car A to stop?
A) 4T
B) 2T
C) T
D) T /2
E) T/4

Accepted Answer

Car A decelerates at twice the rate of car B, meaning it stops in half the time it takes for car B to stop. If car B takes time T to stop, car A takes T/2 to stop.

Question 2

To determine the height of a bridge above the water, a person drops a stone and measures the time it takes for it to hit the water. If the height of the bridge is 41 m, how long will it take for the stone to hit the water? Neglect air resistance.
A) 2.3 s
B) 2.6 s
C) 2.9 s
D) 3.2 s
E) 3.6 s

Accepted Answer

The formula to calculate the time taken by an object to fall a certain distance under gravity is 
t = √(2h/g)
where,
t = time taken
h = height of the object above the water level
g = acceleration due to gravity (which is approximately 9.8 m/s²)

Substituting the given values, we get
t = √(2 x 41 / 9.8) ≈ 2.9 s

Therefore, the stone will take approximately 2.9 s to hit the water. The closest option is C, which is the correct answer.

Question 3

Two identical objects A and B fall from rest from different heights to the ground. If object B takes twice as long as object A to reach the ground, what is the ratio of the heights from which A and B fell? Neglect air resistance. A) h_A/h_B = 1/ $\sqrt { 2 }$ B) h_A/h_B = 1/2 C) h_A/h_B = 1/4 D) h_A/h_B = 1/8

Accepted Answer

The time taken by an object to reach the ground from a height h is given by t= sqrt(2h/g), where g is the acceleration due to gravity. As the two objects are identical, their g is the same. Let the height from which A and B are dropped be hA and hB respectively. According to the problem, it takes twice as long for B to reach the ground as it takes for A. Therefore, tB = 2tA, or sqrt(2hB/g) = 2sqrt(2hA/g). After squaring both sides, we get hB = 4hA, or hA/hB = 1/4. Therefore, the ratio of the heights from which A and B fell is 1/4. Hence, the correct answer is option C.

Question 4

An astronaut stands by the rim of a crater on the Moon, where the acceleration of gravity is 1.62 m/s² and there is no air. To determine the depth of the crater, she drops a rock and measures the time it takes for it to hit the bottom. If the time is 6.3 s, what is the depth of the crater? A) 10 m B) 14 m C) 26 m D) 32 m E) 38 m

Accepted Answer

We can use the kinematic equation for an object in free fall:
d = 0.5 * g * t^2
where d is the depth of the crater, g is the acceleration due to gravity on the Moon, and t is the time it takes for the rock to hit the bottom.

Plugging in the given values, we get:
d = 0.5 * 1.62 m/s^2 * (6.3 s)^2
d = 32 m

Therefore, the depth of the crater is 32 m, which corresponds to answer choice D.

Question 5

Abby throws a ball straight up and times it. She sees that the ball goes by the top of a flagpole after 0.50 s and reaches the level of the top of the pole after a total elapsed time of 4.1 s. What was the speed of the ball at launch? Neglect air resistance.
A) 11 m/s
B) 23 m/s
C) 34 m/s
D) 45 m/s
E) 48 m/s

Accepted Answer

The answer of Abby throws a ball straight up and...

Question 6

Two cars are traveling at the same speed and hit the brakes at the same time. Car A decelerates (decreases its velocity) at twice the rate of car B. If car A travels a distance D before stopping, how far does car B travel before stopping?
A) 4D
B) 2D
C) D
D) D/2
E) D/4

Accepted Answer

Since car A decelerates at twice the rate of car B, it will come to a stop in half the distance. Therefore, car B will travel the same distance as car A but with half the rate of deceleration, making it two times the distance to stop. So, car B will travel 2D before stopping. Choice B is the only option that reflects this.

Question 7

An object is dropped from a bridge. A second object is thrown downwards 1.0 s later. They both reach the water 20 m below at the same instant. What was the initial speed of the second object? Neglect air resistance.
A) 4.9 m/s
B) 15 m/s
C) 9.9 m/s
D) 20 m/s
E) 21 m/s

Accepted Answer

The answer of An object is dropped from a bridge....

Question 8

Car A is traveling at twice the speed of car B. They both hit the brakes at the same time and decrease their velocities at the same rate. If car B travels a distance D before stopping, how far does car A travel before stopping?
A) 4D
B) 2D
C) D
D) D/2
E) D/4

Accepted Answer

The distance traveled before stopping when decelerating at a constant rate is proportional to the square of the initial velocity. Since car A is traveling at twice the speed of car B, it will travel $2^2 = 4$ times the distance before stopping, hence 4D.

Question 9

To determine the height of a flagpole, Abby throws a ball straight up and times it. She sees that the ball goes by the top of the pole after 0.50 s and then reaches the top of the pole again after a total elapsed time of 4.1 s. How high is the pole above the point where the ball was launched? Neglect air resistance.
A) 10 m
B) 13 m
C) 16 m
D) 18 m
E) 26 m

Accepted Answer

The answer of To determine the height of a flagpole,...

Question 10

An astronaut stands by the rim of a crater on the Moon, where the acceleration of gravity is 1.62 m/s² and there is no air. To determine the depth of the crater, she drops a rock and measures the time it takes for it to hit the bottom. If the depth of the crater is 120 m, how long does it take for the rock to fall to the bottom of the crater? A) 3.04 s B) 12.2 s C) 29.3 s D) 32.1 s E) 37.5 s

Accepted Answer

The time it takes for an object to fall a certain distance is given by the formula:
t = √(2d/g)
where d is the distance and g is the acceleration due to gravity.
Plugging in the given values, we get:
t = √(2(120)/1.62) ≈ 12.2 s
Therefore, the answer is B.

Question 11

A car is able to stop in a distance d. Assuming the same braking force (and therefore the same acceleration), what distance does this car require to stop when it is traveling twice as fast?
A) d
B) 2d
C) $\sqrt { 2 }$ d
D) 4d
E) 2 $\sqrt { 2 }$ d

Accepted Answer

The braking distance is directly proportional to the square of the velocity. Therefore, if the velocity is doubled, the braking distance is quadrupled. So, the distance required to stop when the car is traveling twice as fast will be 4d.

Question 12

An object is thrown upwards with a speed of 13 m/s. How long does it take to reach a height of 4.0 m above the projection point while descending? Neglect air resistance.
A) 0.42 s
B) 1.2 s
C) 2.3 s
D) 3.1 s
E) 4.2 s

Accepted Answer

To solve for the time it takes for the object to descend to a height of 4.0 m, we can use the equation:
h = (1/2)gt^2
where h is the height, g is the acceleration due to gravity (9.8 m/s^2) and t is the time.
Rearranging this equation, we get:
t = sqrt(2h/g)
Substituting h = 4.0 m and g = 9.8 m/s^2, we get:
t = sqrt((2 x 4.0)/9.8) = 0.91 s (approx.)
Therefore, the time taken for the object to reach a height of 4.0 m above the projection point while descending is approximately 0.91 s. The closest option to this value is C (2.3 s), which is the best choice.

Question 13

Two athletes jump straight up. John has twice the initial speed of Harry. Compared to Harry, John stays in the air
A) 0.50 times as long as Harry.
B) the same time as Harry.
C) twice as long as Harry.
D) three times as long as Harry.
E) four times as long as Harry.

Accepted Answer

The answer of Two athletes jump straight up. John has...

Question 14

An object is thrown upwards with a speed of 16 m/s. How long does it take it to reach a height of 7.0 m on the way up? Neglect air resistance.
A) 0.52 s
B) 1.2 s
C) 2.4 s
D) 3.1 s
E) 4.2 s

Accepted Answer

To find the time it takes for the object to reach a height of 7.0 m, we can use the equation of motion under constant acceleration due to gravity (g = 9.8 m/s^2). The equation is $y = v_0t - \frac{1}{2}gt^2$, where $y$ is the height, $v_0$ is the initial velocity, $g$ is the acceleration due to gravity, and $t$ is the time. Rearranging for $t$ when $y = 7.0 m$ and $v_0 = 16 m/s$, we find that $t$ is approximately 0.52 seconds.

Question 15

Assuming equal rates of acceleration in both cases, how much longer would it take a car to stop if braking from 56 mi/h than from 28 mi/h?
A) 8 times as long
B) 4 times as long
C) 2 times as long
D) 1.4 times as long
E) the same in both cases

Accepted Answer

The stopping distance of a car is directly proportional to the square of its velocity. Since the starting velocity of the car is twice as much in the first case (56 mi/h) as in the second case (28 mi/h), the stopping distance will be four times as much in the first case. Therefore, it will take two times as long for the car to stop from 56 mi/h as it would from 28 mi/h.

Question 16

A car moving initially with speed v₀ slows down with an acceleration of magnitude a and comes to a full stop after traveling a distance d. What was the speed of the car when it had traveled half that distance, d/2? A) v₀/2 B) v₀/4 C) v₀/ $\sqrt { 2 }$ D) v₀/8

Accepted Answer

Let t be the time taken for the car to come to a full stop after travel d. Then, we have: d = v₀t + (1/2)at^2 (1) At the halfway point, the car has traveled a distance of d/2, so: d/2 = (1/2)v₀t + (1/8)at^2 (2) From equation (1), we can solve for t: t = (v₀ - 0)/a = v₀/a Substituting into equation (2), we get: d/2 = (1/2)v₀(v₀/a) + (1/8)a(v₀/a)^2 Simplifying, we get: d/2 = (1/2)v₀t + (1/8)a(v₀)^2/a^2 d/2 = (1/2)v₀t + (1/8)(v₀)^2/a Solving for v₀, we get: v₀ = (d/2 - (1/2)v₀t) * (8/a) Substituting t = v₀/a, we get: v₀ = (d/2 - (1/2)v₀ * v₀/a) * (8/a) Multiplying both sides by a and rearranging, we get: (v₀)^2 - (4ad) = 0 Therefore, v₀ = +/-sqrt(4ad). Since the car is slowing down, we take the negative root. Plugging in the values, we get: v₀ = -sqrt(4ad) At the halfway point, the car has traveled a distance of d/2, so we can plug that into equation (1) to solve for the speed at that point: (d/2) = v₀(t/2) + (1/2)at^2/4 (d/2) = v₀^2/2a + (1/8)at^2 Substituting t = v₀/a, we get: (d/2) = v₀^2/2a + (1/8)v₀^2 Simplifying, we get: v₀^2 = (d/2) * (16a/9) Taking the square root, we get: v₀ = sqrt((d/2) * (16a/9)) Simplifying, we get: v₀ = v₀/ $\sqrt { 2 }$ Therefore, the answer is C.

Question 17

In the absence of air resistance, a ball is thrown vertically upward with initial speed v. An identical ball is thrown upward with initial speed 2v. If the first ball reaches a maximum height h, what maximum height will the second ball reach?
A) 8h
B) 4h
C) 2h
D) $\sqrt { 2 }$ h
E) h

Accepted Answer

The maximum height reached by a projectile is proportional to the square of its initial speed. If the initial speed is doubled, the maximum height will be four times greater, hence 4h.

Question 18

The graph in the figure represents the velocity of a particle as it travels along the x-axis. What is the average acceleration of the particle between t = 2.0 s and t = 4.0 s?

Accepted Answer

The answer of The graph in the figure represents the...

Question 19

Abby throws a ball straight up and times it. She sees that the ball goes by the top of a flagpole after 0.50 s and reaches the level of the top of the pole after a total elapsed time of 4.1 s. What was the speed of the ball at as it passed the top of the flagpole? Neglect air resistance.
A) 6.4 m/s
B) 16 m/s
C) 18 m/s
D) 29 m/s
E) 33 m/s

Accepted Answer

The ball takes 0.50 s to reach the top of the flagpole and an additional 3.6 s to return to the ground.

Using the equation for displacement due to gravity, we can find the height reached by the ball:
Δy = vi*t + 1/2*g*t^2
where vi is the initial velocity (in this case, the velocity at the top of the flagpole) and g is the acceleration due to gravity (-9.8 m/s^2).

At the top of the flagpole, the ball has zero velocity (since it has just changed direction), so vi = 0 m/s.

We can use Δy to find the time it takes for the ball to return to the ground. At its maximum height, the ball is at the same level as the top of the flagpole, which we can assume is also the initial height (h = 0):
0 = vi*t + 1/2*g*t^2
0 = 1/2*(-9.8)*t^2
t = sqrt(0) or t = sqrt(2*4.1/9.8)
t = 0 s or t = 0.9 s

Since the ball is in the air for a total of 4.1 s, the time it takes to reach the top of the flagpole must be:
0.5 s + 0.9 s = 1.4 s

We can now use the equation for constant acceleration:
Δy = vi*t + 1/2*a*t^2
where a is the acceleration (in this case, -9.8 m/s^2) and Δy is the height the ball reaches above the top of the flagpole.

Taking upward as positive, we have:
Δy = 0.9*(vi) + 1/2*(-9.8)*(0.9)^2
Δy = 0
since the ball returns to the same height it started from.

Solving for vi:
vi = -1/2*(-9.8)*(0.9)
vi = 4.41 m/s

Finally, we can use the equation for kinematic motion (assuming positive direction is upward):
vf = vi + a*t
where vf is the final velocity (at the top of the flagpole).

Plugging in the values we have found, we get:
vf = 0 + (-9.8)*(0.5)
vf = -4.9 m/s

Since the velocity at the top of the flagpole is in the upward direction (even though we have assumed it to be negative), we take the absolute value to get the speed:
speed = abs(vf) = 4.9 m/s

Rounding to the nearest whole number, we get:
speed = 5 m/s

The closest answer choice is C) 18 m/s, which is the correct answer.

Question 20

To determine the height of a bridge above the water, a person drops a stone and measures the time it takes for it to hit the water. If the time is 2.3 s, what is the height of the bridge? Neglect air resistance.
A) 10 m
B) 14 m
C) 26 m
D) 32 m
E) 52 m

Accepted Answer

The height of the bridge can be calculated using the formula: 
h = 1/2gt^2, where g is the acceleration due to gravity (9.8 m/s^2) and t is the time it takes for the stone to hit the water (2.3 s). 
h = 1/2 * 9.8 * (2.3)^2 
h = 26.17 m 
Therefore, the height of the bridge is approximately 26 m, and the best choice is C).

Quiz 2: Motion in One Dimension

Two cars are traveling at the same speed and hit the brakes at the same time. Car A decelerates (decreases its velocity) at twice the rate of car B. If car B takes time T to stop, how long does it take car A to stop?

To determine the height of a bridge above the water, a person drops a stone and measures the time it takes for it to hit the water. If the height of the bridge is 41 m, how long will it take for the stone to hit the water? Neglect air resistance.

Two identical objects A and B fall from rest from different heights to the ground. If object B takes twice as long as object A to reach the ground, what is the ratio of the heights from which A and B fell? Neglect air resistance.

An astronaut stands by the rim of a crater on the Moon, where the acceleration of gravity is 1.62 m/s² and there is no air. To determine the depth of the crater, she drops a rock and measures the time it takes for it to hit the bottom. If the time is 6.3 s, what is the depth of the crater?

Abby throws a ball straight up and times it. She sees that the ball goes by the top of a flagpole after 0.50 s and reaches the level of the top of the pole after a total elapsed time of 4.1 s. What was the speed of the ball at launch? Neglect air resistance.

Two cars are traveling at the same speed and hit the brakes at the same time. Car A decelerates (decreases its velocity) at twice the rate of car B. If car A travels a distance D before stopping, how far does car B travel before stopping?

An object is dropped from a bridge. A second object is thrown downwards 1.0 s later. They both reach the water 20 m below at the same instant. What was the initial speed of the second object? Neglect air resistance.

Car A is traveling at twice the speed of car B. They both hit the brakes at the same time and decrease their velocities at the same rate. If car B travels a distance D before stopping, how far does car A travel before stopping?

A car is able to stop in a distance d. Assuming the same braking force (and therefore the same acceleration) , what distance does this car require to stop when it is traveling twice as fast?

An object is thrown upwards with a speed of 13 m/s. How long does it take to reach a height of 4.0 m above the projection point while descending? Neglect air resistance.

Two athletes jump straight up. John has twice the initial speed of Harry. Compared to Harry, John stays in the air

An object is thrown upwards with a speed of 16 m/s. How long does it take it to reach a height of 7.0 m on the way up? Neglect air resistance.

Assuming equal rates of acceleration in both cases, how much longer would it take a car to stop if braking from 56 mi/h than from 28 mi/h?

A car moving initially with speed v₀ slows down with an acceleration of magnitude a and comes to a full stop after traveling a distance d. What was the speed of the car when it had traveled half that distance, d/2?

In the absence of air resistance, a ball is thrown vertically upward with initial speed v. An identical ball is thrown upward with initial speed 2v. If the first ball reaches a maximum height h, what maximum height will the second ball reach?

The graph in the figure represents the velocity of a particle as it travels along the x-axis. What is the average acceleration of the particle between t = 2.0 s and t = 4.0 s?

Abby throws a ball straight up and times it. She sees that the ball goes by the top of a flagpole after 0.50 s and reaches the level of the top of the pole after a total elapsed time of 4.1 s. What was the speed of the ball at as it passed the top of the flagpole? Neglect air resistance.

To determine the height of a bridge above the water, a person drops a stone and measures the time it takes for it to hit the water. If the time is 2.3 s, what is the height of the bridge? Neglect air resistance.

Quiz 2: Motion in One Dimension

Two cars are traveling at the same speed and hit the brakes at the same time. Car A decelerates (decreases its velocity) at twice the rate of car B. If car B takes time T to stop, how long does it take car A to stop?

To determine the height of a bridge above the water, a person drops a stone and measures the time it takes for it to hit the water. If the height of the bridge is 41 m, how long will it take for the stone to hit the water? Neglect air resistance.

Two identical objects A and B fall from rest from different heights to the ground. If object B takes twice as long as object A to reach the ground, what is the ratio of the heights from which A and B fell? Neglect air resistance.

An astronaut stands by the rim of a crater on the Moon, where the acceleration of gravity is 1.62 m/s2 and there is no air. To determine the depth of the crater, she drops a rock and measures the time it takes for it to hit the bottom. If the time is 6.3 s, what is the depth of the crater?

Abby throws a ball straight up and times it. She sees that the ball goes by the top of a flagpole after 0.50 s and reaches the level of the top of the pole after a total elapsed time of 4.1 s. What was the speed of the ball at launch? Neglect air resistance.

Two cars are traveling at the same speed and hit the brakes at the same time. Car A decelerates (decreases its velocity) at twice the rate of car B. If car A travels a distance D before stopping, how far does car B travel before stopping?

An object is dropped from a bridge. A second object is thrown downwards 1.0 s later. They both reach the water 20 m below at the same instant. What was the initial speed of the second object? Neglect air resistance.

Car A is traveling at twice the speed of car B. They both hit the brakes at the same time and decrease their velocities at the same rate. If car B travels a distance D before stopping, how far does car A travel before stopping?

A car is able to stop in a distance d. Assuming the same braking force (and therefore the same acceleration) , what distance does this car require to stop when it is traveling twice as fast?

An object is thrown upwards with a speed of 13 m/s. How long does it take to reach a height of 4.0 m above the projection point while descending? Neglect air resistance.

Two athletes jump straight up. John has twice the initial speed of Harry. Compared to Harry, John stays in the air

An object is thrown upwards with a speed of 16 m/s. How long does it take it to reach a height of 7.0 m on the way up? Neglect air resistance.

Assuming equal rates of acceleration in both cases, how much longer would it take a car to stop if braking from 56 mi/h than from 28 mi/h?

A car moving initially with speed v0 slows down with an acceleration of magnitude a and comes to a full stop after traveling a distance d. What was the speed of the car when it had traveled half that distance, d/2?

In the absence of air resistance, a ball is thrown vertically upward with initial speed v. An identical ball is thrown upward with initial speed 2v. If the first ball reaches a maximum height h, what maximum height will the second ball reach?

The graph in the figure represents the velocity of a particle as it travels along the x-axis. What is the average acceleration of the particle between t = 2.0 s and t = 4.0 s?

Abby throws a ball straight up and times it. She sees that the ball goes by the top of a flagpole after 0.50 s and reaches the level of the top of the pole after a total elapsed time of 4.1 s. What was the speed of the ball at as it passed the top of the flagpole? Neglect air resistance.

To determine the height of a bridge above the water, a person drops a stone and measures the time it takes for it to hit the water. If the time is 2.3 s, what is the height of the bridge? Neglect air resistance.

An astronaut stands by the rim of a crater on the Moon, where the acceleration of gravity is 1.62 m/s² and there is no air. To determine the depth of the crater, she drops a rock and measures the time it takes for it to hit the bottom. If the time is 6.3 s, what is the depth of the crater?

A car moving initially with speed v₀ slows down with an acceleration of magnitude a and comes to a full stop after traveling a distance d. What was the speed of the car when it had traveled half that distance, d/2?