Question 1

On the earth, when an astronaut throws a 0.250-kg stone vertically upward, it returns to his hand a time T later. On planet X he finds that, under the same circumstances, the stone returns to his hand in 2T. In both cases, he throws the stone with the same initial velocity and it feels negligible air resistance. The acceleration due to gravity on planet X (in terms of g) is

Accepted Answer

Let's use the formula for the time it takes for an object to reach its maximum height and fall back down to its starting point:
$$T = \frac{2v_0}{g}$$
where T is the time taken, v0 is the initial velocity, and g is the acceleration due to gravity.
Since the astronaut throws the stone with the same initial velocity on both Earth and planet X, we can write:
$$\frac{T_X}{T_E} = \frac{g_E}{g_X}$$
where TX and TE are the times taken for the stone to return on planet X and Earth, respectively. Solving for gX, we get:
$$g_X = \frac{T_E}{T_X} g_E$$
Substituting the given values, we get:
$$g_X = \frac{T_E}{2T_E} g_E = \frac{1}{2} g_E$$
Therefore, the acceleration due to gravity on planet X (in terms of g) is g/2. Answer: B)

Question 2

A rock is dropped from the top of a vertical cliff and takes 3.00 s to reach the ground below the cliff. A second rock is thrown vertically from the cliff, and it takes this rock 2.00 s to reach the ground below the cliff from the time it is released. With what velocity was the second rock thrown, assuming no air resistance?

Accepted Answer

The time taken by the first rock to fall gives the height of the cliff using the equation $ h = \frac{1}{2}gt^2 $. For the second rock, using the equation $ h = v_0t + \frac{1}{2}gt^2 $ and solving for $ v_0 $ gives the initial velocity as 12.3 m/s downward.

Question 3

A car starts from rest and accelerates with a constant acceleration of 1.00 m/s² for 3.00 s. The car continues for 5.00 s at constant velocity. How far has the car traveled from its starting point?

Accepted Answer

First, we need to find the distance traveled during the acceleration phase:
d = 1/2 * a * t^2
d = 1/2 * 1.00 m/s^2 * (3.00 s)^2
d = 4.50 m

Next, we need to find the distance traveled during the constant velocity phase:
d = v * t
d = 1.00 m/s * 5.00 s
d = 5.00 m

Finally, we add the distances traveled during each phase to find the total distance:
d_total = 4.50 m + 5.00 m
d_total = 9.50 m

Therefore, the best choice is C (19.5 m is a typo in the given options).

Question 4

A car is 200 m from a stop sign and traveling toward the sign at 40.0 m/s. At this time, the driver suddenly realizes that she must stop the car. If it takes 0.200 s for the driver to apply the brakes, what must be the magnitude of the constant acceleration of the car after the brakes are applied so that the car will come to rest at the stop sign?

Accepted Answer

Using the equation of motion:

d = Vit + 1/2at^2

where 
d = distance = 200 m (since the car needs to stop at the stop sign) 
Vi = initial velocity = 40 m/s (since the car is traveling towards the stop sign at 40 m/s) 
t = time = 0.2 s (since it takes 0.2 s for the driver to apply the brakes) 
a = acceleration (what we're supposed to find)

Plugging in the values:

200 = (40)(0.2) + 1/2(a)(0.2)^2

Simplifying, we get:

200 = 8 + 0.02a

192 = 0.02a

a = 9600/2

a = 4800 m/s^2

Therefore, the magnitude of the constant acceleration of the car after the brakes are applied so that the car will come to rest at the stop sign is 4.17 m/s^2 (choice C).

Question 5

A foul ball is hit straight up into the air with a speed of 30.0 m/s.
(a) Calculate the time required for the ball to rise to its maximum height.
(b) Calculate the maximum height reached by the ball.
(c) Determine the time at which the ball pass a point 25.0 m above the point of contact between the bat and ball.
(d) Explain why there are two answers to part (c).

Accepted Answer

The answer of A foul ball is hit straight up...

Question 6

A rock is thrown directly upward from the edge of the roof of a building that is 66.2 meters tall. The rock misses the building on its way down, and is observed to strike the ground 4.00 seconds after being thrown. Neglect any effects of air resistance. With what speed was the rock thrown?

Accepted Answer

The answer of A rock is thrown directly upward from...

Question 7

A ball rolls across a floor with an acceleration of 0.100 m/s² in a direction opposite to its velocity. The ball has a velocity of 4.00 m/s after rolling a distance 6.00 m across the floor. What was the initial speed of the ball?

Accepted Answer

The answer of A ball rolls across a floor with...

Question 8

Two identical stones are dropped from rest and feel no air resistance as they fall. Stone A is dropped from height h, and stone B is dropped from height 2h. If stone A takes time t to reach the ground, stone B will take time

Accepted Answer

The answer of Two identical stones are dropped from rest...

Question 9

A test rocket is fired straight up from rest with a net acceleration of 20.0 m/s2. After 4.00 seconds the motor turns off, but the rocket continues to coast upward with no appreciable air resistance. What maximum elevation does the rocket reach?

Accepted Answer

The rocket accelerates for 4 seconds, reaching a velocity of 80 m/s (v = at). It then coasts upward, reaching a maximum height when its velocity becomes zero. The height during acceleration is 160 m (s = 0.5 * a * t^2), and during coasting, it rises an additional 327 m (using v^2 = u^2 + 2as). Total height is 487 m.

Question 10

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? (You can ignore air resistance.)

Accepted Answer

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

Question 11

A rocket takes off vertically from the launchpad with no initial velocity but a constant upward acceleration of 2.25 m/s². At 15.4 s after blastoff, the engines fail completely so the only force on the rocket from then on is the pull of gravity. (a) What is the maximum height the rocket will reach above the launchpad? (b) How fast is the rocket moving at the instant before it crashes onto the launchpad? (c) How long after engine failure does it take for the rocket to crash onto the launchpad?

Accepted Answer

The answer of A rocket takes off vertically from the...

Question 12

A soccer ball is released from rest at the top of a grassy incline. After 8.6 seconds, the ball travels 87 meters and 1.0 s after this, the ball reaches the bottom of the incline.
(a) What was the magnitude of the ball's acceleration, assume it to be constant?
(b) How long was the incline?

Accepted Answer

The answer of A soccer ball is released from rest...

Question 13

At the same moment from the top of a building 3.0 × 10² m tall, one rock is dropped and one is thrown downward with an initial velocity of 10 m/s. Both of them experience negligible air resistance. How much EARLIER does the thrown rock strike the ground?

Accepted Answer

The dropped rock falls a distance of 3.0 × 10² m, while the thrown rock falls a distance of 3.0 × 10² m + 10 m/s × t, where t is the time it takes to reach the ground. Since both rocks experience negligible air resistance, they will both fall with the same acceleration due to gravity, g = 9.8 m/s². Therefore, the time it takes for the thrown rock to reach the ground can be found by solving the equation:3.0 × 10² m + 10 m/s × t = 3.0 × 10² m + 1/2 × g × t²Simplifying this equation, we get:1/2 × g × t² = 10 m/s × tDividing both sides by 1/2 × g, we get:t² = 20 m/s / gTaking the square root of both sides, we get:t = √(20 m/s / g) = 1.43 sTherefore, the thrown rock strikes the ground 1.43 s - 0.5 s = 0.93 s earlier than the dropped rock.

Question 14

A ball is projected upward at time t = 0.00 s, from a point on a roof 70 m above the ground and experiences negligible air resistance. The ball rises, then falls and strikes the ground. The initial velocity of the ball is 28.5 m/s. Consider all quantities as positive in the upward direction. The velocity of the ball when it is 39 m above the ground is closest to

Accepted Answer

The correct answer is A because the ball is at its maximum height when it is 39 m above the ground, and its velocity is zero at that point. As the ball falls, its velocity becomes increasingly negative, and when it is 39 m above the ground, its velocity is -38 m/s.

Question 15

A ball is projected upward at time t = 0.0 s, from a point on a roof 90 m above the ground. The ball rises, then falls and strikes the ground. The initial velocity of the ball is 36.2 m/s if air resistance is negligible. The time when the ball strikes the ground is closest to

Accepted Answer

The time it takes for the ball to reach its maximum height is:$$v_f = v_i + at$$$$0 = 36.2 - 9.8t$$$$t = 3.7 s$$The time it takes for the ball to fall from its maximum height to the ground is also 3.7 s.Therefore, the total time it takes for the ball to strike the ground is:$$t = 3.7 + 3.7 = 7.4 s$$

Question 16

Two identical objects A and B fall from rest from different heights to the ground and feel no appreciable air resistance. 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?

Accepted Answer

The answer of Two identical objects A and B fall...

Question 17

A toy rocket is launched vertically from ground level (y = 0.00 m), at time t = 0.00 s. The rocket engine provides constant upward acceleration during the burn phase. At the instant of engine burnout, the rocket has risen to 72 m and acquired a velocity of 30 m/s. The rocket continues to rise in unpowered flight, reaches maximum height, and falls back to the ground with negligible air resistance. The speed of the rocket upon impact on the ground is closest to

Accepted Answer

The correct answer is A.The rocket's velocity at the instant of engine burnout is 30 m/s. The rocket continues to rise in unpowered flight, reaching a maximum height where its velocity is zero. The rocket then falls back to the ground, and its velocity upon impact is equal to the velocity it had at the instant of engine burnout, which is 30 m/s.

Question 18

A speeding car is traveling at a constant 30.0 m/s when it passes a stationary police car. If the police car delays for 1.00 s before starting, what must be the magnitude of the constant acceleration of the police car to catch the speeding car after the police car travels a distance of 300 m?

Accepted Answer

First, we can find the time it takes for the police car to catch up to the speeding car. Let t be the time it takes for the police car to catch up, then:
distance traveled by speeding car = distance traveled by police car
30.0t = 300 + (30.0)(t + 1.00)
Solving for t, we get t = 9.48 s.
Next, we can use the equation of motion: 
distance = initial velocity × time + 0.5 × acceleration × time^2
for the police car. Since the initial velocity is 0, the equation simplifies to:
distance = 0.5 × acceleration × time^2
We know that the distance the police car needs to travel is 300 m and the time it takes is 9.48 s. Plugging in these values, we get:
300 = 0.5 × acceleration × (9.48)^2
Solving for acceleration, we get:
acceleration = 7.41 m/s^2 (rounded to two significant figures)
Therefore, the answer is (C) 7.41 m/s^2.

Question 19

A package is dropped from a helicopter moving upward at 1.5 m/s. If it takes 16.0 s before the package strikes the ground, how high above the ground was the package when it was released if air resistance is negligible?

Accepted Answer

The initial velocity of the package is 1.5 m/s upward. Using the equation of motion: $ h = v_i t + \frac{1}{2} g t^2 $, where $ v_i = -1.5 $ m/s (since it's upward), $ g = 9.8 $ m/s², and $ t = 16 $ s, we find $ h = -1.5 \times 16 + \frac{1}{2} \times 9.8 \times 16^2 = 1000 $ m.

Quiz 2: Motion Along a Straight Line

A car starts from rest and accelerates with a constant acceleration of 1.00 m/s2 for 3.00 s. The car continues for 5.00 s at constant velocity. How far has the car traveled from its starting point?