Linear Motion - AP Physics B

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Question

If a 15kg ball takes five seconds to strike the ground when released from rest, at what height was the ball dropped?

Answer

Using the equation we can find the distance at which the ball was dropped. Notice that the mass of the ball does not matter in this problem. We are told that the ball is dropped from rest making, , thus we have . When we plug in our values, and assuming that acceleration is equal to gravity (10m/s2) we find that = 125m.

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Question

How far will an object travel after ten seconds if it is dropped into a bottomless pit?

Answer

Since the object is dropped, the inital velocity is zero. Gravity is the only acceleration, the time is ten seconds, and the distance at which the object travels is unknown.

The equation can be used to find the distance traveled.

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Question

How long does it take an object to travel a distance of 30m from rest at a constant acceleration of 2m/s2?

Answer

Using the equation , we can solve for time.

Since the object started at rest, . Now we are left with the equation .

$30m=\frac{1}{2}(2\frac{m}{s^2})t^2$

Plugging in the remaining values we can find that t = 5.5s.

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Question

You are driving at a speed of $20\frac{m}{s}$ and suddenly, a tree falls down on the road blocking your path. You slam on your brakes to avoid hitting the fallen tree and thus, come to a complete stop. You were at a distance of away from the tree when you hit the brakes. Assuming that your vehicle does not skid, what is the minimum deceleration needed to avoid hitting the fallen tree?

Answer

Use the following kinematic equation where the initial velocity is , final velocity is , and the distance traveled is $\Delta x$ .

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

We can use the values in the question to solve for the acceleration.

$(0\frac{m}{s})^2=(20\frac{m}{s})^2+2a(28m)$

We rearrange the equation to solve for the acceleration.

$a=\frac{0^2-(20\frac{m}{s})^2}{2(28\ \text{m})}$

$a=-7.14\ \frac{\text{m}}{\text{s}^2}$

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Question

A person on top of a tall building drops a rock. How long will it take for the rock to reach the ground? Ignore air resistance.

Answer

Use the following kinematic equation: $y=y_o+v_ot+\frac{1}{2}at^2$ .

We can choose the ground to be the zero distance so that and $y_o=200\ \text{m}$ .

Also, the initial speed is zero.

The kinematic equation simplifies using these values.

$0=y_o+\frac{1}{2}at^2$

Solve to isolate the time variable.

$t=\sqrt{\frac{-2y_o}{a}}$

We know that the acceleration is the acceleration due to gravity. Now we can plug in the known values and solve.

$t=\sqrt{\frac{-2(200\ \text{m})}{-9.8\ \frac{\text{m}}{\text{s}^2}}}$

$t=\sqrt{40.8s^2}=6.4s$

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Question

A block is placed at a height of on a ramp at an incline of . It slides down from rest with an acceleration of $2.2\frac{m}{s^2}$ due to a frictional force. What is the velocity of the block when it reaches the bottom of incline?

Answer

To find the final velocity we use the following kinematic equation.

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

The block starts from rest, so . The incline is eighteen meters high, so we can find the distance (hypotenuse) as follows:

$\Delta x=\frac{18m}{sin24^\circ}$

$\Delta x = 44.25m$

We know that the acceleration is $a=2.2\frac{m}{s^2}$ .

Plug in the values to find the velocity at the bottom of the incline.

$v_f^2=(0\frac{m}{s})^2+2(2.2\frac{m}{s^2})(44.25m)$

$v_f=\sqrt{194.7\frac{m^2}{s^2}}=14\frac{m}{s}$

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Question

A is dropped from a height of . A picture is taken when the ball is from the ground with an exposure time of . If the actual diameter of the ball is , what will the vertical diameter of the ball appear to be in the picture?

Answer

The first step to solving this problem will be to find the velocity of the ball at the point when the picture is taken. We know the initial velocity of the ball (zero), the displacement, and the acceleration. Using the appropriate kinematics formula, we can solve for the final velocity.

$v_f^2=v_o^2+2a\Delta y$

$v_f^2=(0\frac{m}{s})^2+2(-9.8\frac{m}{s^2})(5m-20m)$

$v_f^2=294\frac{m^2}{s^2}$

$v_f=17.15\frac{m}{s}$

Now that we know how fast the ball is traveling when the picture is taken, we can find the distance it travels while the shutter is open. This distance will become a motion blur, making the vertical diameter of the ball appear stretched.

$v=\frac{d}{t}\rightarrow d=vt$

$d=(17.15\frac{m}{s})(0.005s)=0.086m$

During the exposure period, the ball will travel , or . The diameter of the ball in the vertical direction will appear to be distorted by this distance.

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Question

A person stands on the edge of a straight -high cliff and holds a ball over the edge. The person tosses the ball directly upward with an initial speed of $5 \frac{m}{s}$ . How long will it take the ball to hit the ground at the base of the cliff, below?

Assume $10 \frac{m}{s^2}$ for gravitational acceleration.

Answer

This is projectile motion in the vertical direction only, subject to the equation of motion: $(x-x_o) = v_ot + \frac{1}{2}at^2$ .

For this discussion, one can define the downward direction as negative. For projectile motion, $a = -10 \frac{m}{s^2}$ (gravitational acceleration, or ).

In this case, the ball ends up below where is started, so .

The initial velocity, , is $5 \frac{m}{s}$ (upward, thus positive).

With all this, the projectile motion equation becomes:

$- 20 m = (5 \frac{m}{s})t - \frac{1}{2}(10 \frac{m}{s^2})t^2$

This can be solved for using the quadratic formula:

$t = \frac{-b \pm \sqrt{b^{2}-4ac)}) }{2a}$

The result is:

$t = \frac{5\pm 20.62}{10} = 2.562 s$ (or ).

Only the positive answer option is physically possible, and is thus our correct answer.

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Question

If a 15kg ball takes five seconds to strike the ground when released from rest, at what height was the ball dropped?

Answer

Using the equation we can find the distance at which the ball was dropped. Notice that the mass of the ball does not matter in this problem. We are told that the ball is dropped from rest making, , thus we have . When we plug in our values, and assuming that acceleration is equal to gravity (10m/s2) we find that = 125m.

Compare your answer with the correct one above

Question

How far will an object travel after ten seconds if it is dropped into a bottomless pit?

Answer

Since the object is dropped, the inital velocity is zero. Gravity is the only acceleration, the time is ten seconds, and the distance at which the object travels is unknown.

The equation can be used to find the distance traveled.

Compare your answer with the correct one above

Question

How long does it take an object to travel a distance of 30m from rest at a constant acceleration of 2m/s2?

Answer

Using the equation , we can solve for time.

Since the object started at rest, . Now we are left with the equation .

$30m=\frac{1}{2}(2\frac{m}{s^2})t^2$

Plugging in the remaining values we can find that t = 5.5s.

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Question

You are driving at a speed of $20\frac{m}{s}$ and suddenly, a tree falls down on the road blocking your path. You slam on your brakes to avoid hitting the fallen tree and thus, come to a complete stop. You were at a distance of away from the tree when you hit the brakes. Assuming that your vehicle does not skid, what is the minimum deceleration needed to avoid hitting the fallen tree?

Answer

Use the following kinematic equation where the initial velocity is , final velocity is , and the distance traveled is $\Delta x$ .

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

We can use the values in the question to solve for the acceleration.

$(0\frac{m}{s})^2=(20\frac{m}{s})^2+2a(28m)$

We rearrange the equation to solve for the acceleration.

$a=\frac{0^2-(20\frac{m}{s})^2}{2(28\ \text{m})}$

$a=-7.14\ \frac{\text{m}}{\text{s}^2}$

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Question

A person on top of a tall building drops a rock. How long will it take for the rock to reach the ground? Ignore air resistance.

Answer

Use the following kinematic equation: $y=y_o+v_ot+\frac{1}{2}at^2$ .

We can choose the ground to be the zero distance so that and $y_o=200\ \text{m}$ .

Also, the initial speed is zero.

The kinematic equation simplifies using these values.

$0=y_o+\frac{1}{2}at^2$

Solve to isolate the time variable.

$t=\sqrt{\frac{-2y_o}{a}}$

We know that the acceleration is the acceleration due to gravity. Now we can plug in the known values and solve.

$t=\sqrt{\frac{-2(200\ \text{m})}{-9.8\ \frac{\text{m}}{\text{s}^2}}}$

$t=\sqrt{40.8s^2}=6.4s$

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Question

A block is placed at a height of on a ramp at an incline of . It slides down from rest with an acceleration of $2.2\frac{m}{s^2}$ due to a frictional force. What is the velocity of the block when it reaches the bottom of incline?

Answer

To find the final velocity we use the following kinematic equation.

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

The block starts from rest, so . The incline is eighteen meters high, so we can find the distance (hypotenuse) as follows:

$\Delta x=\frac{18m}{sin24^\circ}$

$\Delta x = 44.25m$

We know that the acceleration is $a=2.2\frac{m}{s^2}$ .

Plug in the values to find the velocity at the bottom of the incline.

$v_f^2=(0\frac{m}{s})^2+2(2.2\frac{m}{s^2})(44.25m)$

$v_f=\sqrt{194.7\frac{m^2}{s^2}}=14\frac{m}{s}$

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Question

A is dropped from a height of . A picture is taken when the ball is from the ground with an exposure time of . If the actual diameter of the ball is , what will the vertical diameter of the ball appear to be in the picture?

Answer

The first step to solving this problem will be to find the velocity of the ball at the point when the picture is taken. We know the initial velocity of the ball (zero), the displacement, and the acceleration. Using the appropriate kinematics formula, we can solve for the final velocity.

$v_f^2=v_o^2+2a\Delta y$

$v_f^2=(0\frac{m}{s})^2+2(-9.8\frac{m}{s^2})(5m-20m)$

$v_f^2=294\frac{m^2}{s^2}$

$v_f=17.15\frac{m}{s}$

Now that we know how fast the ball is traveling when the picture is taken, we can find the distance it travels while the shutter is open. This distance will become a motion blur, making the vertical diameter of the ball appear stretched.

$v=\frac{d}{t}\rightarrow d=vt$

$d=(17.15\frac{m}{s})(0.005s)=0.086m$

During the exposure period, the ball will travel , or . The diameter of the ball in the vertical direction will appear to be distorted by this distance.

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Question

A person stands on the edge of a straight -high cliff and holds a ball over the edge. The person tosses the ball directly upward with an initial speed of $5 \frac{m}{s}$ . How long will it take the ball to hit the ground at the base of the cliff, below?

Assume $10 \frac{m}{s^2}$ for gravitational acceleration.

Answer

This is projectile motion in the vertical direction only, subject to the equation of motion: $(x-x_o) = v_ot + \frac{1}{2}at^2$ .

For this discussion, one can define the downward direction as negative. For projectile motion, $a = -10 \frac{m}{s^2}$ (gravitational acceleration, or ).

In this case, the ball ends up below where is started, so .

The initial velocity, , is $5 \frac{m}{s}$ (upward, thus positive).

With all this, the projectile motion equation becomes:

$- 20 m = (5 \frac{m}{s})t - \frac{1}{2}(10 \frac{m}{s^2})t^2$

This can be solved for using the quadratic formula:

$t = \frac{-b \pm \sqrt{b^{2}-4ac)}) }{2a}$

The result is:

$t = \frac{5\pm 20.62}{10} = 2.562 s$ (or ).

Only the positive answer option is physically possible, and is thus our correct answer.

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Question

An object is shot from the ground at 75m/s at an angle of 45⁰ above the horizontal. How high does the object get before beginning its descent?

Answer

The velocity must be broken down into x (horizontal) and y (vertical) components. We can use the y component to find how high the object gets. To find vertical velocity, vy, use $v_y = v_o sin\theta$ .

$v_y = (75 \frac{m}{s})sin 45^o = 53 \frac{m}{s}$

Next we find how long it takes to reach the top of its trajectory using .

$0 \frac{m}{s} = 53 \frac{m}{s} + (-10 \frac{m}{s^2})t$

t = 5.3s

Finally, find how high the object goes with $d = v_ot +\frac{1}{2}at^{2}$ .

$d = (53\frac{m}{s})5.3s + \frac{1}{2}(-10\frac{m}{s^2})(5.3s)^{2} = 140 m$

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Question

An object is shot from the ground at 125m/s at an angle of 30o above the horizontal. How far away does the object land?

Answer

First, find the horizontal (x) and vertical (y) components of the velocity

$v_x = v_o cos\theta$

$v_x = (125 \frac{m}{s}) cos(30^o) = 108\frac{m}{s}$

$v_y = v_o sin\theta$

$v_y = (125\frac{m}{s})sin(30^o) = 62.5\frac{m}{s}$

Next, find how long the object is in the air by calculating the time it takes it to reach the top of its path, and doubling that number.

$0 \frac{m}{s} = 62.5 \frac{m}{s} + (-10 \frac{m}{s^2})t$

t = 6.25s

Total time in the air is therefore 12.5s (twice this value).

Finally, find distance traveled my multiplying horizontal velocity and time.

$d_x = 108 \frac{m}{s} (12.5s) = 1,350 m$

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Question

A 2kg box is at the top of a frictionless ramp at an angle of . The top of the ramp is 30m above the ground. The box is sitting still while at the top of the ramp, and is then released.

When the box is released, how long will it take the box to reach the ground?

Answer

We can start this problem by determining how far the box will travel on the ramp before hitting the ground. Since the vertical distance is 30m and we know the angle of the ramp, we can determine the length of the hypotenuse using the equation $\small cos(60^{\circ}) = \frac{30}{x}$ .

$\frac{1}{2}=\frac{30m}{x}\rightarrow x=60m$

Now that we have the distance traveled, we can determine the acceleration on the box due to gravity. Because the box is on a sloped surface, the box will not experience the full acceleration of gravity, but will instead be accelerated at a value of $\small gsin\Theta$ . Since the angle is 60o, the acceleration on the box is $\small 8.66 \frac{m}{s^{2}}.$

$gsin\Theta=(10)sin(60)=8.66\frac{m}{s^2}$

Finally, we can plug these values into the following distance equation and solve for time.

$\small x = x_{0} - \frac{1}{2}(gsin\Theta )t^{2}$

$\small 0 = 60 - \frac{1}{2}(8.66)t^{2}$

$\small \small t = 3.72 s$

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Question

A 2kg box is at the top of a frictionless ramp at an angle of 60o. The top of the ramp is 30m above the ground. The box is sitting still while at the top of the ramp, and is then released.

What is the velocity of the box just before it hits the ground?

Answer

We can start this problem by determining how much time it takes the box the reach the ground. Since the vertical distance is 30m and we know the angle of the ramp.

$\frac{1}{2}=\frac{30m}{x}\rightarrow x=60m$

Now that we have the distance traveled, we can determine the acceleration on the box due to gravity, using the equation $\small gsin\Theta$ .

$gsin\Theta=(10)sin(60)=8.66\frac{m}{s^2}$

We can plug these values into the following distance equation and solve for time.

$\small x = x_{0} - \frac{1}{2}(gsin\Theta )t^{2}$

$\small 0 = 60 - \frac{1}{2}(8.66)t^{2}$

$\small \small t = 3.72 s$

Now that we know the acceleration on the box and the time of travel, we can use the equation $\small v = v_{0} + at$ to solve for the velocity.

$\small v = 0 + (8.66\frac{m}{s^{2}})(3.72 s) = 32.2 \frac{m}{s}$

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