Question 1

A stone is thrown with an initial upward velocity of 7.0 m/s.If we take upward as the positive direction,what is the velocity of the stone after 0.50 s,assuming there is no air?

Accepted Answer

The velocity of the stone after 0.50 s can be calculated using the equation $v = u + at$, where $v$ is the final velocity, $u$ is the initial velocity (7.0 m/s upwards), $a$ is the acceleration due to gravity (-9.8 m/s$^2$ downwards), and $t$ is the time (0.50 s). Substituting the values, we get $v = 7.0 m/s + (-9.8 m/s$^2$) \times 0.50 s = 7.0 m/s - 4.9 m/s = 2.1 m/s$ upwards.

Question 2

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 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?

Accepted Answer

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

Question 4

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/ $\mathrm { s }^2$ How long after car A begins to accelerate does it take car A to overtake car B?

Accepted Answer

To find the time it takes for car A to overtake car B, we set up equations for their positions as functions of time and solve for when they are equal. Car A's position is given by $x_A = 22t + 0.5 \times 2.4 \times t^2$, and car B's position is $x_B = 29t + 300$. Setting $x_A = x_B$ and solving for $t$, we find $t \approx 19.0$ seconds.

Question 5

A hammer is thrown upward with a speed of 14 m/s on the surface of planet X where the acceleration of object in free fall is 3.5 m/s² and there is no atmosphere.What is the speed of the hammer after 8.0 s?

Accepted Answer

The answer of A hammer is thrown upward with a...

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?

Accepted Answer

The answer of A baseball is hit with a bat...

Question 7

A rock is thrown upward with a speed of 15 m/s on the surface of planet X where the acceleration of objects in free fall is 2.5 m/s² and there is no atmosphere.How long does it take for the rock to return to where it was thrown?

Accepted Answer

The answer of A rock is thrown upward with a...

Question 8

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

Accepted Answer

To find the time to reach maximum height, use the formula $t = \frac{v}{g}$, where $v = 12 \, \text{m/s}$ is the initial velocity and $g = 1.5 \, \text{m/s}^2$ is the acceleration due to gravity. Plugging in the values, $t = \frac{12}{1.5} = 8 \, \text{s}$.

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 when it is in the state of free fall.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?

Accepted Answer

The average acceleration is calculated using the formula $a = \frac{\Delta v}{\Delta t} = \frac{29 \, \mathrm{m/s}}{0.15 \, \mathrm{s}} = 193.33 \, \mathrm{m/s^2}$. To express this in g's, divide by $9.8 \, \mathrm{m/s^2}$: $ \frac{193.33}{9.8} \approx 19.74 $, which rounds to 20 g.

Question 10

Human reaction time is usually greater than 0.10 s.If your friend holds a ruler between your fingers and releases it without warning,how far can you expect the ruler to fall before you catch it,assuming that air has no effect on the ruler?

Accepted Answer

The ruler falls due to gravity, so we use the formula: 
d = 1/2*g*t^2 
where d is the distance fallen, g is the acceleration due to gravity (9.8 m/s^2), and t is the reaction time (assumed to be 0.10 s). 
Plugging in the values, we get: 
d = 1/2 * 9.8 m/s^2 * (0.10 s)^2 
d = 0.049 m = 4.9 cm 
Therefore, the ruler will fall at least 4.9 cm before it is caught.

Question 11

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

Accepted Answer

First, we can find the time it takes the rock to reach its maximum height using the kinematic equation: 
h = ut + (1/2)at^2 
where h =  maximum height, u = initial velocity (12 m/s), a = acceleration due to gravity on planet X (-1.5 m/s^2), and t = time at maximum height. 
Setting h = 0 (maximum height), we get: 
0 = 12t - (1/2)(1.5)t^2 
Simplifying and solving for t, we get: 
t = 8 seconds 
Next, we can use the same kinematic equation to find the maximum height: 
h = ut + (1/2)at^2 
Plugging in the values we know, we get: 
h = 12(8) + (1/2)(-1.5)(8)^2 
Simplifying, we get: 
h = 48 meters 
Therefore, the answer is C, 48 meters.

Question 12

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?

Accepted Answer

We can use the equation:

distance = initial velocity * time + 1/2 * acceleration * time^2

Since acceleration is constant, we can use:

average velocity = (final velocity + initial velocity)/2

to find the average velocity in the time interval from 5.0 s to 8.0 s:

average velocity = (12.0 m/s + 8.0 m/s)/2 = 10.0 m/s

Using this, we can find the distance traveled:

distance = (8.0 m/s)(3.0 s) + 1/2(acceleration)(3.0 s)^2

distance = 24.0 m + 4.5a

We don't know the value of acceleration, but we can eliminate it by using another equation:

final velocity^2 = initial velocity^2 + 2 * acceleration * distance

We can solve for distance:

distance = (final velocity^2 - initial velocity^2)/(2 * acceleration)

Plugging in the values we know:

distance = (12.0 m/s)^2 - (8.0 m/s)^2)/(2 * acceleration)

distance = 20.0 m/acceleration

Now we can equate the two expressions for distance and solve for acceleration:

24.0 m + 4.5a = 20.0 m/acceleration

4.5a^2 + 20.0 = 0

a = -2.666 m/s^2

The negative sign means that the acceleration is in the opposite direction of the motion. We can now use the equation for distance and plug in the value of acceleration:

distance = (12.0 m/s)^2 - (8.0 m/s)^2)/(2 * (-2.666 m/s^2))

distance = 30 m

Therefore, the distance traveled in the time interval from 5.0 s to 8.0 s is 30 m. So, the answer is C.

Question 13

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?

Accepted Answer

Using the kinematic equation $x = vt + \frac{1}{2}at^2$, where $x$ is the distance travelled, $v$ is the initial velocity (which is 0 m/s), $a$ is the acceleration ($6.00	ext{ m/s}^2$), and $t$ is the time interval (3.00 s), we get:

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

Therefore, the car travels 27.0 m in the first 3.00 s. The answer is (C).

Question 14

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 of freely falling objects on the planet?

Accepted Answer

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

Question 15

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?

Accepted Answer

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

Question 16

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 m/s² until it is brought to a halt.The distance traveled by the train while slowing down is closest to

Accepted Answer

Explanation:The train travels 2.1 km while accelerating, 9.6 km while moving at a constant velocity, and 3.8 km while decelerating. The total distance traveled is 2.1 km + 9.6 km + 3.8 km = 15.5 km.

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?

Accepted Answer

The initial velocity of the car, u = 32.0 m/s
The uniform acceleration, a = -3.50 m/s^2 (negative since it is opposing the direction of motion)
The final velocity of the car, v = 0 (since the car comes to a stop)

We can use the following kinematic equation to find the distance traveled by the car:

v^2 = u^2 + 2as

0 = (32.0 m/s)^2 + 2(-3.50 m/s^2)s

Solving for s, we get:

s = 146 m

Therefore, the car travels a distance of 146 m after the driver puts on the brakes until it comes to a stop. The answer is (D).

Question 18

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?

Accepted Answer

We can use the equation: 
v^2 = u^2 + 2as 
where
v = final velocity (0 m/s)
u = initial velocity (unknown)
a = acceleration (-3.5 m/s^2) 
s = distance (-30.0 m)
Plugging in the values, we get: 
0 = u^2 - 2(3.5)(-30.0) 
Simplifying, we get: 
u = 14.5 m/s 
Therefore, the car's original speed was 14.5 m/s.

Question 19

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?

Accepted Answer

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

Question 20

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?

Accepted Answer

We can solve this problem using the kinematic equation:

v^2 = u^2 + 2as

Where: 
v = final velocity (0 m/s) 
u = initial velocity 
a = acceleration 
s = displacement (120 m)

We can rearrange this equation to solve for u:

u = sqrt(v^2 - 2as)

u = sqrt((26.0 m/s)^2 - 2(−a)(120 m))

We know that the acceleration is negative because the car is slowing down:

a = - v^2 / 2s
a = - (26.0 m/s)^2 / (2 x 120 m)

a = -11.05 m/s^2

Substituting this value into the previous equation:

u = sqrt((26.0 m/s)^2 - 2(-11.05 m/s^2)(120 m))

u = 18.4 m/s

Therefore, the car was traveling at 18.4 m/s when it was 60.0 m past the point where the brakes were applied. The answer is B.

Quiz 2: Kinematics: Motion in One Dimension

A stone is thrown with an initial upward velocity of 7.0 m/s.If we take upward as the positive direction,what is the velocity of the stone after 0.50 s,assuming there is no air?

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?

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/ s2\mathrm { s }^2s2 How long after car A begins to accelerate does it take car A to overtake car B?

A hammer is thrown upward with a speed of 14 m/s on the surface of planet X where the acceleration of object in free fall is 3.5 m/s2 and there is no atmosphere.What is the speed of the hammer after 8.0 s?

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 rock is thrown upward with a speed of 15 m/s on the surface of planet X where the acceleration of objects in free fall is 2.5 m/s2 and there is no atmosphere.How long does it take for the rock to return to where it was thrown?

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