Question 1

Assuming equal rates of uniform acceleration in both cases, how much further would you travel if braking from $56 \mathrm { mi } / \mathrm { h }$ to rest than from $28 \mathrm { mi } / \mathrm { h }$ ?
A) 4 times farther
B) 3.2 times farther
C) 4.8 times farther
D) 5.2 times farther

Accepted Answer

The stopping distance is proportional to the square of the initial speed. Therefore, if the speed is doubled from $28 \mathrm{mi/h}$ to $56 \mathrm{mi/h}$, the stopping distance will be $2^2 = 4$ times greater.

Question 2

A car accelerates from $5.0 \mathrm {~m} / \mathrm { s }$ to $21 \mathrm {~m} / \mathrm { s }$ at a constant rate of $3.0 \mathrm {~m} / \mathrm { s } ^ { 2 }$ How far does it travel while accelerating?
A) 69 m
B) 207 m
C) 41 m
D) 117 m

Accepted Answer

The car's initial velocity is 5.0 m/s, its final velocity is 21 m/s, and its acceleration is 3.0 m/s^2. We can use the following equation to calculate the distance traveled while accelerating:```distance = (final velocity^2 - initial velocity^2) / (2 * acceleration)```Plugging in the values we know, we get:```distance = (21 m/s)^2 - (5.0 m/s)^2) / (2 * 3.0 m/s^2) = 69 m```Therefore, the car travels 69 m while accelerating.

Question 3

A train starts from rest and accelerates uniformly until it has traveled 5.6 km and acquired a forward velocity of $42 \mathrm {~m} / \mathrm { s }$ The train then moves at a constant velocity of $42 \mathrm {~m} / \mathrm { s }$ for 420 s. The train then slows down uniformly at $0.065 \mathrm {~m} / \mathrm { s } ^ { 2 }$ until it is brought to a halt. The acceleration during the first 5.6 km of travel is closest to which of the following?
A) 0.16 m/ $s ^ { 2 }$
B) 0.14 m/ $s ^ { 2 }$
C) 0.17 m/ $s ^ { 2 }$
D) 0.19 m/ $s ^ { 2 }$
E) 0.20 m/ $s ^ { 2 }$

Accepted Answer

Using the equation $v^2 = u^2 + 2as$, where $v = 42 \, \text{m/s}$, $u = 0$, and $s = 5600 \, \text{m}$, we solve for $a$ to find $a \approx 0.16 \, \text{m/s}^2$.

Question 4

An astronaut on a strange new planet having no atmosphere finds that she can jump up to a maximum height of 27 m when her initial upward speed is 6.0 m/s. What is the magnitude of the acceleration due to gravity on the planet?

Accepted Answer

The answer of An astronaut on a strange new planet...

Question 5

A laser is thrown upward with a speed of 12 m/s on the surface of planet X where the acceleration due to gravity is 1.5 m/s² and there is no atmosphere. How long does it take for the laser to reach the maximum height? A) 8.0 s B) 11 s C) 14 s D) 16 s

Accepted Answer

Using the equation for maximum height:

h = (v^2) / (2g)

where:
v = initial velocity (12 m/s)
g = acceleration due to gravity (1.5 m/s^2)

Substituting the values:

h = (12^2) / (2 x 1.5) = 48 m

Using the equation for time of flight:

t = 2v/g

Substituting the values:

t = 2 x 12 / 1.5 = 16 s

But this gives us the total time taken for the laser to go up and come back down.

Therefore, the time taken for the laser to reach the maximum height is:

t/2 = 16/2 = 8 s

So, the best choice is A) 8.0 s.

Question 6

A baseball is hit with a bat and, as a result, its direction is completely reversed and its speed is doubled. If the actual contact with the bat lasts 0.45 s, what is the ratio of the magnitude of the average acceleration of the ball to its original speed? A) 6.7 s^-¹ B) 4.4 s^-1 C) 2.2 s^-1 D) 0.15 s^-1

Accepted Answer

The change in velocity is three times the original speed (since the direction is reversed and speed is doubled). The average acceleration is the change in velocity divided by the time of contact, so the ratio is 3/0.45 = 6.7 s^-1.

Question 7

A car starts from rest and accelerates at a steady 6.00 m/s². How far does it travel in the first 3.00 s? A) 9.00 m B) 18.0 m C) 27.0 m D) 36.0 m E) 54.0 m

Accepted Answer

Using the kinematic equation:

$s = ut + \frac{1}{2}at^2$

Where,

s = distance traveled = ?
u = initial velocity = 0 (car starts from rest)
a = acceleration = 6.00 m/s^2
t = time taken = 3.00 s

Substituting the values, we get:

$s = 0 + \frac{1}{2}(6.00)(3.00)^2 = 27.0 	ext{ m}$

Therefore, the car travels a distance of 27.0 m in the first 3.00 s. The correct answer is option C).

Question 8

Car A is traveling at 22.0 m/s and car B at 29.0 m/s. Car A is 300 m behind car B when the driver of car A accelerates his car with a uniform forward acceleration of 2.40 m/ $s ^ { 2 }$ . How long after car A begins to accelerate does it take car A to overtake car B?
A) 5.50 s
B) 12.6 s
C) 19.0 s
D) 316 s
E) Car A never overtakes car B.

Accepted Answer

To find the time it takes for car A to overtake car B, we set the distance traveled by both cars equal. Using the equations of motion, we solve for time and find it to be approximately 19.0 seconds.

Question 9

Acceleration is sometimes expressed in multiples of g, where $g = 9.8 \mathrm {~m} / \mathrm { s } ^ { 2 }$ is the acceleration of an object due to the earth's gravity. In a car crash, the car's forward velocity may go from $29 \mathrm {~m} / \mathrm { s }$ to $0 \mathrm {~m} / \mathrm { s }$ in $0.15 \mathrm {~s}$ How many g's are experienced, on average, by the driver?
A) 20 g
B) 14 g
C) 24 g
D) 26 g

Accepted Answer

The average acceleration is:$$a = \frac{\Delta v}{\Delta t} = \frac{0 \mathrm {~m} / \mathrm { s } - 29 \mathrm {~m} / \mathrm { s }}{0.15 \mathrm {~s}} = -193.33 \mathrm {~m} / \mathrm { s } ^ { 2 }$$The acceleration in g's is:$$a_g = \frac{a}{g} = \frac{-193.33 \mathrm {~m} / \mathrm { s } ^ { 2 }}{9.8 \mathrm {~m} / \mathrm { s } ^ { 2 }} = -19.73$$The average acceleration experienced by the driver is -19.73 g's, which is closest to 20 g.

Question 10

A soccer ball is released from rest at the top of a grassy incline. After 6.4 seconds the ball has rolled 91 m with constant acceleration, and 1.0 s later it reaches the bottom of the incline.
(a) What was the ball's acceleration?
(b) How long was the incline?

Accepted Answer

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

Question 11

A car is traveling with a constant speed of 30.0 m/s when the driver suddenly applies the brakes, causing the car to slow down with a constant acceleration. The car comes to a stop in a distance of 120 m. What was the acceleration of the car as it slowed down? A) 3.75 m/s² B) 4.00 m/s² C) 4.25 m/s² D) 4.50 m/s² E) 4.75 m/s²

Accepted Answer

The acceleration can be found using the equation $v^2 = u^2 + 2as$, where $v = 0$ (final velocity), $u = 30.0$ m/s (initial velocity), $s = 120$ m (distance), and $a$ is the acceleration. Rearranging for $a$, we get $a = -\frac{u^2}{2s}$. Substituting the given values, $a = -\frac{(30.0)^2}{2 \times 120} = -\frac{900}{240} = -3.75$ m/s$^2$. The negative sign indicates deceleration.

Question 12

A train starts from rest and accelerates uniformly until it has traveled 2.1 km and acquired a forward velocity of $ 24 \mathrm{~m} / \mathrm{s} $ The train then moves at a constant velocity of $ 24 \mathrm{~m} / \mathrm{s} $ for 400 s. The train then slows down uniformly at $ 0.065 \mathrm{~m} / \mathrm{s}^{2} $ until it is brought to a halt. The distance traveled by the train while slowing down is closest to
A) 4.4 km.
B) 4.2 km.
C) 4.0 km.
D) 3.8 km.
E) 3.6 km.

Accepted Answer

The distance traveled while slowing down can be calculated using the formula $ v^2 = u^2 + 2as $, where $ v = 0 $, $ u = 24 \, \text{m/s} $, and $ a = -0.065 \, \text{m/s}^2 $. Solving for $ s $ gives approximately 4.4 km.

Question 13

An airplane starts from rest and accelerates at a constant 10.8 m/s². What is its speed at the end of a 400 m-long runway? A) 37.0 m/s B) 93.0 m/s C) 65.7 m/s D) 4320 m/s E) 186 m/s

Accepted Answer

The answer of An airplane starts from rest and accelerates...

Question 14

A car is traveling with a constant speed when the driver suddenly applies the brakes, causing the car to slow down with a constant acceleration of magnitude 3.50 m/s². If the car comes to a stop in a distance of 30.0 m, what was the car's original speed? A) 10.2 m/s B) 14.5 m/s C) 105 m/s D) 210 m/s E) 315 m/s

Accepted Answer

We can use the kinematic equation: 
vf^2 = vi^2 + 2ad 
where vf = final velocity (0 m/s since the car comes to a stop), vi = initial velocity (what we're trying to find), a = acceleration (-3.50 m/s^2 since the car slows down), and d = distance (30.0 m). 
Plugging in the known values, we get: 
0^2 = vi^2 + 2(-3.50)(30.0) 
Solving for vi, we get: 
vi = 14.5 m/s 
Therefore, the car's original speed was 14.5 m/s, which is answer choice B.

Question 15

A car is traveling at 26.0 m/s when the driver suddenly applies the brakes, causing the car to slow down with constant acceleration. The car comes to a stop in a distance of 120 m. How fast was the car moving when it was 60.0 m past the point where the brakes were applied?
A) 22.5 m/s
B) 18.4 m/s
C) 15.0 m/s
D) 12.1 m/s
E) 9.20 m/s

Accepted Answer

Using the equations of motion, we can solve for the acceleration of the car:

vf^2 = vi^2 + 2ad

0^2 = 26.0^2 + 2a(120)

a = -5.40 m/s^2

Now we can use another equation of motion to solve for the velocity of the car when it was 60 m past the point where the brakes were applied:

vf^2 = vi^2 + 2ad

vf^2 = vi^2 + 2(-5.40)(60)

vf^2 = vi^2 - 648

vi^2 = vf^2 + 648

vi = sqrt(vf^2 + 648)

vi = sqrt((0.5(26.0))^2 + 648)

vi = 18.4 m/s

Therefore, the answer is B) 18.4 m/s.

Question 16

An airplane needs to reach a forward velocity of $203.0 \mathrm {~km} / \mathrm { h }$ to take off. On a $2000 - \mathrm { m }$ runway, what is the minimum uniform acceleration necessary for the plane to take flight if it starts from rest? A) 0.79 m/s² B) 0.87 m/s² C) 0.95 m/s² D) 1.0 m/s²

Accepted Answer

To find the minimum uniform acceleration, use the equation $v^2 = u^2 + 2as$, where $v = 203.0 \, \text{km/h} = 56.39 \, \text{m/s}$, $u = 0$, and $s = 2000 \, \text{m}$. Solving for $a$, we get $a = \frac{v^2}{2s} = \frac{(56.39)^2}{2 \times 2000} \approx 0.79 \, \text{m/s}^2$.

Question 17

A car is moving with a speed of 32.0 m/s. The driver sees an accident ahead and slams on the brakes, causing the car to slow down with a uniform acceleration of magnitude 3.50 m/s². How far does the car travel after the driver put on the brakes until it comes to a stop? A) 4.57 m B) 9.14 m C) 112 m D) 146 m E) 292 m

Accepted Answer

We can use the kinematic equation:

$v_f^2 = v_i^2 + 2a\Delta x$

where $v_f$ is the final velocity (0 m/s), $v_i$ is the initial velocity (32.0 m/s), $a$ is the acceleration (-3.50 m/s^2), and $\Delta x$ is the distance traveled.

Rearranging and solving for $\Delta x$, we get:

$\Delta x = \frac{v_f^2 - v_i^2}{2a} = \frac{0^2 - (32.0 	ext{ m/s})^2}{2(-3.50 	ext{ m/s}^2)} \approx 146 	ext{ m}$

Therefore, the answer is D, 146 m.

Question 18

A laser is thrown upward with a speed of 12 m/s on the surface of planet X where the acceleration due to gravity is 1.5 m/s² and there is no atmosphere. What is the maximum height reached by the laser? A) 8.0 m B) 18 m C) 48 m D) 144 m

Accepted Answer

We can use the kinematic equation: 
vf^2 = vi^2+2ad 
where 
vf = 0 (at maximum height)
vi = 12 m/s
a = -1.5 m/s^2 (negative because it's going against gravity)
d = max height we want to find

Solving for d: 
0^2 = (12 m/s)^2 + 2(-1.5 m/s^2)d 
d = 48 m

Therefore, the maximum height reached by the laser is 48 meters.

Question 19

A stone is thrown with an initial upward velocity of 7.0 m/s and experiences negligible air resistance. If we take upward as the positive direction, what is the velocity of the stone after 0.50 s?
A) 2.1 m/s
B) 4.9 m/s
C) -2.1 m/s
D) -4.9 m/s
E) 0.00 m/s

Accepted Answer

Using the equation v = u + at, where v is final velocity, u is initial velocity, a is acceleration, and t is time, we get:

v = u + at
v = 7.0 m/s - 9.8 m/s^2 * 0.5 s  (since the acceleration due to gravity is downwards)
v ≈ 2.1 m/s

Since the stone is moving upwards, the velocity is positive. Therefore, the answer is A) 2.1 m/s.

Question 20

A car is moving with a constant acceleration. At time t = 5.0 s its velocity is 8.0 m/s in the forward direction, and at time t = 8.0 s its velocity is 12.0 m/s forward. What is the distance traveled in that interval of time?
A) 10 m
B) 20 m
C) 30 m
D) 40 m
E) 50 m

Accepted Answer

We can use the equation: 
distance = initial velocity x time + 1/2 x acceleration x time^2

We are given: 
initial velocity = 8.0 m/s 
final velocity = 12.0 m/s 
time interval = 3.0 s

We need to find acceleration first: 
acceleration = (final velocity - initial velocity)/time interval 
acceleration = (12.0 m/s - 8.0 m/s)/3.0 s 
acceleration = 1.33 m/s^2 (rounded to two decimal places)

Now we can use the above equation to find the distance: 
distance = 8.0 m/s x 3.0 s + 1/2 x 1.33 m/s^2 x (3.0 s)^2 
distance = 24.0 m + 6.0 m 
distance = 30.0 m

Therefore, the best choice is (C).

Quiz 2: Motion in One Dimension

Assuming equal rates of uniform acceleration in both cases, how much further would you travel if braking from 56mi/h56 \mathrm { mi } / \mathrm { h }56mi/h to rest than from 28mi/h28 \mathrm { mi } / \mathrm { h }28mi/h ?

A car accelerates from 5.0 m/s5.0 \mathrm {~m} / \mathrm { s }5.0 m/s to 21 m/s21 \mathrm {~m} / \mathrm { s }21 m/s at a constant rate of 3.0 m/s23.0 \mathrm {~m} / \mathrm { s } ^ { 2 }3.0 m/s2 How far does it travel while accelerating?

An astronaut on a strange new planet having no atmosphere finds that she can jump up to a maximum height of 27 m when her initial upward speed is 6.0 m/s. What is the magnitude of the acceleration due to gravity on the planet?

A laser is thrown upward with a speed of 12 m/s on the surface of planet X where the acceleration due to gravity is 1.5 m/s2 and there is no atmosphere. How long does it take for the laser to reach the maximum height?

A baseball is hit with a bat and, as a result, its direction is completely reversed and its speed is doubled. If the actual contact with the bat lasts 0.45 s, what is the ratio of the magnitude of the average acceleration of the ball to its original speed?

A car starts from rest and accelerates at a steady 6.00 m/s2. How far does it travel in the first 3.00 s?

Car A is traveling at 22.0 m/s and car B at 29.0 m/s. Car A is 300 m behind car B when the driver of car A accelerates his car with a uniform forward acceleration of 2.40 m/ s2s ^ { 2 }s2 . How long after car A begins to accelerate does it take car A to overtake car B?

A soccer ball is released from rest at the top of a grassy incline. After 6.4 seconds the ball has rolled 91 m with constant acceleration, and 1.0 s later it reaches the bottom of the incline. (a) What was the ball's acceleration? (b) How long was the incline?

A car is traveling with a constant speed of 30.0 m/s when the driver suddenly applies the brakes, causing the car to slow down with a constant acceleration. The car comes to a stop in a distance of 120 m. What was the acceleration of the car as it slowed down?

An airplane starts from rest and accelerates at a constant 10.8 m/s2. What is its speed at the end of a 400 m-long runway?

A car is traveling with a constant speed when the driver suddenly applies the brakes, causing the car to slow down with a constant acceleration of magnitude 3.50 m/s2. If the car comes to a stop in a distance of 30.0 m, what was the car's original speed?

A car is traveling at 26.0 m/s when the driver suddenly applies the brakes, causing the car to slow down with constant acceleration. The car comes to a stop in a distance of 120 m. How fast was the car moving when it was 60.0 m past the point where the brakes were applied?

An airplane needs to reach a forward velocity of 203.0 km/h203.0 \mathrm {~km} / \mathrm { h }203.0 km/h to take off. On a 2000−m2000 - \mathrm { m }2000−m runway, what is the minimum uniform acceleration necessary for the plane to take flight if it starts from rest?

A car is moving with a speed of 32.0 m/s. The driver sees an accident ahead and slams on the brakes, causing the car to slow down with a uniform acceleration of magnitude 3.50 m/s2. How far does the car travel after the driver put on the brakes until it comes to a stop?

A laser is thrown upward with a speed of 12 m/s on the surface of planet X where the acceleration due to gravity is 1.5 m/s2 and there is no atmosphere. What is the maximum height reached by the laser?

A stone is thrown with an initial upward velocity of 7.0 m/s and experiences negligible air resistance. If we take upward as the positive direction, what is the velocity of the stone after 0.50 s?

A car is moving with a constant acceleration. At time t = 5.0 s its velocity is 8.0 m/s in the forward direction, and at time t = 8.0 s its velocity is 12.0 m/s forward. What is the distance traveled in that interval of time?

Assuming equal rates of uniform acceleration in both cases, how much further would you travel if braking from $56 \mathrm { mi } / \mathrm { h }$ to rest than from $28 \mathrm { mi } / \mathrm { h }$ ?

A car accelerates from $5.0 \mathrm {~m} / \mathrm { s }$ to $21 \mathrm {~m} / \mathrm { s }$ at a constant rate of $3.0 \mathrm {~m} / \mathrm { s } ^ { 2 }$ How far does it travel while accelerating?

A laser is thrown upward with a speed of 12 m/s on the surface of planet X where the acceleration due to gravity is 1.5 m/s² and there is no atmosphere. How long does it take for the laser to reach the maximum height?

A car starts from rest and accelerates at a steady 6.00 m/s². How far does it travel in the first 3.00 s?

Car A is traveling at 22.0 m/s and car B at 29.0 m/s. Car A is 300 m behind car B when the driver of car A accelerates his car with a uniform forward acceleration of 2.40 m/ $s ^ { 2 }$ . How long after car A begins to accelerate does it take car A to overtake car B?

An airplane starts from rest and accelerates at a constant 10.8 m/s². What is its speed at the end of a 400 m-long runway?

A car is traveling with a constant speed when the driver suddenly applies the brakes, causing the car to slow down with a constant acceleration of magnitude 3.50 m/s². If the car comes to a stop in a distance of 30.0 m, what was the car's original speed?

An airplane needs to reach a forward velocity of $203.0 \mathrm {~km} / \mathrm { h }$ to take off. On a $2000 - \mathrm { m }$ runway, what is the minimum uniform acceleration necessary for the plane to take flight if it starts from rest?

A car is moving with a speed of 32.0 m/s. The driver sees an accident ahead and slams on the brakes, causing the car to slow down with a uniform acceleration of magnitude 3.50 m/s². How far does the car travel after the driver put on the brakes until it comes to a stop?

A laser is thrown upward with a speed of 12 m/s on the surface of planet X where the acceleration due to gravity is 1.5 m/s² and there is no atmosphere. What is the maximum height reached by the laser?