Question 1

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

Firstly, find the time it takes for the speeding car to travel 300m:

distance = speed x time

time = distance/speed = 300/30 = 10s

Now we know that the police car needs to travel the same distance of 300m in 10 - 1 = 9s, as the speeding car started 1s earlier. Using the same formula, we can find the required acceleration of the police car:

distance = 0.5 x acceleration x time^2

300 = 0.5 x acceleration x 9^2

acceleration = 7.41 m/s^2

Question 2

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 using the equation: 
$d = \frac{1}{2}at^2 = \frac{1}{2} (1.00	ext{ m/s}^2)(3.00	ext{ s})^2 = 4.50	ext{ m}$

Next, we find the distance traveled during the constant velocity phase using the equation: 
$d = vt = (1.00	ext{ m/s})(5.00	ext{ s}) = 5.00	ext{ m}$

Finally, we add the distances from both phases: 
$4.50	ext{ m} + 5.00	ext{ m} = 9.50	ext{ m}$

Therefore, the best choice is C: 19.5 m since the car traveled a total distance of 9.50 m, not 24.0 m, 19.5 m, 4.50 m, or 15.0 m.

Question 3

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 time it takes for an object to fall a distance h from rest is given by the formula: 
t = sqrt(2h/g), where g is the acceleration due to gravity (9.8 m/s^2).

For the dropped rock: 
t = sqrt(2*3.0 × 10^12/9.8) = 244.9 seconds

For the thrown rock, the initial velocity is 10 m/s downwards, so its motion is given by:
h = vt + (1/2)at^2
where v is the initial velocity, a is the acceleration due to gravity (-9.8 m/s^2), and t is the time elapsed.

Solving for t:
3.0 × 10^12 = 10t + (1/2)(-9.8)t^2
7.84t^2 - 10t - 3.0 × 10^12 = 0
Using the quadratic formula, t = 344.9 seconds or -441.7 seconds. Since the time can't be negative, we discard that solution.

Thus, the thrown rock takes 344.9 seconds to hit the ground, while the dropped rock takes 244.9 seconds to hit the ground. The difference in time is:
344.9 - 244.9 = 100 seconds = 0.95 seconds (to two significant figures)

Therefore, the thrown rock hits the ground 0.95 seconds earlier than the dropped rock. Therefore, the answer is (A).

Question 4

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 answer of A ball is projected upward at time...

Question 5

A package is dropped from a helicopter moving upward at 15 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 answer of A package is dropped from a helicopter...

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 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 8

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 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

To find the maximum elevation, we first calculate the distance traveled during the powered phase and then add the distance it coasts upward after the motor turns off. 1. During the powered phase (first 4 seconds), we use the equation $d = \frac{1}{2}at^2$, where $a = 20.0 \, m/s^2$ and $t = 4.0 \, s$. This gives us $d = \frac{1}{2} \times 20.0 \times (4.0)^2 = 160 \, m$.2. To find the velocity at the end of the powered phase, we use $v = at$, giving us $v = 20.0 \times 4.0 = 80.0 \, m/s$.3. For the coasting phase, we use the equation for motion under gravity $v^2 = u^2 + 2as$, where $v = 0$ (at the maximum height), $u = 80.0 \, m/s$, $a = -9.8 \, m/s^2$ (acceleration due to gravity acting downwards), and $s$ is the distance. Rearranging for $s$, we get $s = \frac{u^2}{2g} = \frac{(80.0)^2}{2 \times 9.8} \approx 327 \, m$.4. The total elevation is the sum of the distances in the powered and coasting phases: $160 \, m + 327 \, m = 487 \, m$.Therefore, the maximum elevation reached by the rocket is 487 meters.

Question 10

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 answer of A rock is dropped from the top...

Question 11

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 12

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 answer of A ball is projected upward at time...

Question 13

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 14

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 15

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

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Question 16

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, $x=x_0+v_0t+\frac{1}{2}at^2$, where $x_0$ = 200 m, $v_0$ = 40.0 m/s, $t$ = 0.200 s, and $x$ = 0 m,

$0=200+(40.0)(0.200)+\frac{1}{2}a(0.200)^2$

Solving for $a$, we get:

$a=-\frac{(40.0)(0.200)}{(0.200)^2}= -400.0 \frac{m}{s^2}$

The negative sign indicates that the acceleration is in the direction opposite to the motion of the car, which is the direction needed to bring the car to a stop. The magnitude of the acceleration is $\lvert-400.0\rvert= 400.0 \frac{m}{s^2}$, which is closest to choice C (4.17 m/s$^2$). Therefore, the answer is C.

Question 17

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 18

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

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Question 19

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

The time it takes for the stone to return to the astronaut's hand is given by the equation:
$$T=2\frac{v_0}{g}$$
where $v_0$ is the initial velocity and $g$ is the acceleration due to gravity. Since the initial velocity and stone are the same in both cases, we can equate the expressions for $T$ on both Earth and planet X:
$$T_{	ext{Earth}} = T_{	ext{X}} \Rightarrow 2\frac{v_0}{g_{	ext{Earth}}} = 2\frac{v_0}{g_{	ext{X}}}$$
Simplifying, we get:
$$g_{	ext{X}} = \frac{1}{2} g_{	ext{Earth}}$$
So the acceleration due to gravity on planet X is half that of Earth, which is answer choice B.

Quiz 2: Kinematics in One Dimension

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?

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?

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?