Linear Motion - High School Physics

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Question

A rocket rises vertically from rest, with an acceleration of until it runs out of fuel at an altitude of . After this point, its acceleration is that of gravity downward. How long (total) is it in the air from the time it is launched to when it lands back on the ground?

Answer

This is a multi-step problem and involves looking at the rocket at several points.

Point 0 - launch from the ground

Point 1 - when the fuel runs out

Point 2 - the height that the rocket reaches

Point 3 - when the rocket reaches the ground

Therefore all knowns and unknowns will be denoted by these different points.

To find the total time, it will be necessary to determine the time traveled under the acceleration, the time time traveled without acceleration to the peak and the time from the top of the peak to the ground.

Point 0 to Point 1

$a_{0-1}= 2.8m/s^2$

Equations:

$\Delta x = v_0 \Delta t+\frac{1}{2}a\Delta t^2$

The initial velocity is zero which simplifies the equation to

$\Delta x = \frac{1}{2}a\Delta t^2$

Rearrange to solve for time

$\Delta t =\sqrt{\frac{2\Delta x}{a}}$

$\Delta t =\sqrt{\frac{2(800m-0m)}{(2.8m/s^2)}}$

$\Delta t = 23.9s$

$\Delta t = t_1-t_0$

It will also be important to know the speed that the rocket is traveling at when it is done accelerating.

$v_f = v_0 +a\Delta t$

Point 1 to Point 2

$a_{1-3}= -9.8m/s^2$

Equations:

$v_f=v_0 +a\Delta t$

Rearrange to solve for time

$\Delta t =\frac{v_f-v_0}{a}$

$\Delta t =\frac{0m/s-66.92m/s}{-9.8m/s^2}$

$\Delta t = 6.83s$

$\Delta t = t_2-t_1$

This time is the total time it takes to accelerate then reach the highest point of the peak.

It will also be important to know how high the rocket goes.

$\Delta x = v_0 \Delta t+\frac{1}{2}a\Delta t^2$

$\Delta x = 66.92m/s (6.83s)+\frac{1}{2}(-9.8m/s^2)(6.83s)^2$

$\Delta x = 228.48m$

$\Delta x = x_2-x_1$

This is the total height that it takes the rocket to reach the highest point of the peak.

Point 2 to Point 3

$a_{1-3}= -9.8m/s^2$

Equations:

$\Delta x = v_0 \Delta t+\frac{1}{2}a\Delta t^2$

The initial velocity is zero which simplifies the equation to

$\Delta x = \frac{1}{2}a\Delta t^2$

Rearrange to solve for time

$\Delta t =\sqrt{\frac{2\Delta x}{a}}$

$\Delta t =\sqrt{\frac{2(0m-1028.48m)}{(-9.8m/s^2)}}$

$\Delta t = 14.49s$

$\Delta t = t_3-t_2$

This is the total time for the rocket to go up and then come back down.

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Question

A falling stone takes to travel past a window tall. From what height above the top of the window did the stone fall?

Answer

This is a multi-step problem. The first part to determine is how fast the stone was falling when it passed the window. Knowing this will help us determine the height from which it was originally dropped.

Known:

$\Delta t = 0.29s$

$\Delta x_{window} = -1.8m$

Unknown:

$\Delta x_{building}= ?$

$v_{when it reaches the top of the window} = ?$

Equation:

Using a kinematic equation, determine the speed the rock was moving when it first was at the top of the window.

$\Delta x = v_0 \Delta t+\frac{1}{2}a\Delta t^2$

Rearrange for the initial velocity.

$\frac{\Delta x - \frac{1}{2}a\Delta t^2}{\Delta t}= v_0$

$\frac{-1.8m - \frac{1}{2}(-9.8m/s^2)(.29s)^2}{(0.29)}= v_0$

Using this information as a final velocity it is possible to determine the height from which the stone originally fell. Additionally, since the object is assumed to fall from rest, the initial velocity is .

Knowns:

Unknowns:

$\Delta x = ??$

Equations:

$v_f^2=v_0^2 +2a\Delta x$

$\Delta x =\frac{v_f^2-v_0^2}{2a}$

$\Delta x =\frac{(-4.79m/s)^2- (0m/s)^2}{2(-9.8m/s^2)}$

$\Delta x = -1.17m$

This means that the stone dropped before hitting the window.

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Question

A rock is dropped from a sea cliff, and the sound of it striking the ocean is heard later. If the speed of sound is , how high is the cliff?

Answer

Knowns

$\Delta t_{total} = 3.6s$

$v_{sound}=340m/s$

$v_{0 stone}=0m/s$

Unknowns:

$\Delta x = ?$

Time can be broken up as well. There is a time that it takes the stone to land in the water below. There is also a different time for the sound to reach the person’s ear. This adds up to the total time provided in the problem.

$\Delta t_{rock}= ?$

$\Delta t_{sound} = ?$

Equation:

The stone travels with accelerated motion and the sound travels at a constant velocity.

Each step must be taken into account as the stone travels down the cliff and as the sound travels back.

For the sound traveling the equation required is

$v=\frac{\Delta x}{\Delta t}$

Rearrange the equation to solve for the position as this is one of the factors that will connect both parts of this problem.

$\Delta x = v\Delta t$

For the stone the best equation to be used is

$\Delta x = v_0 \Delta t+\frac{1}{2}a\Delta t^2$

Remember that the stone falls with an initial velocity of 0m/s so the equation can be simplified.

$\Delta x = \frac{1}{2}a\Delta t^2$

These equations are inverses of each other. (One is travelling from the top of the cliff to the ground and the other travels the other direction) So we can set them equal to the inverse of one another.

$\frac{1}{2}a\Delta t_{rock}^2 = -( v\Delta t_{sound})$

At this point, it is important to remember that there is a relationship between the time it takes for the rock to fall and the time for the sound to return to the person’s ear.

$3.6s = \Delta t_{total} =t_{sound} +t_{rock}$

Therefore there is a relationship

$t_{sound}= 3.6s -t_{rock}$

Substitute this relationship into the equation above.

$\frac{1}{2}a\Delta t_{rock}^2 =-( v_{sound}(3.6s -t_{rock}))$

It is now possible to distribute on the right side.

$\frac{1}{2}a\Delta t_{rock}^2 = (-3.6s)v_{sound} +v_{sound}t_{rock}$

This is a quadratic equation. The next step is to rearrange, substitute values and solve the quadratic formula.

$\frac{1}{2}a\Delta t_{rock}^2-v_{sound}t_{rock}+ (3.6s)v_{sound} =0$

$\frac{1}{2}(-9.8m/s^2)\Delta t_{rock}^2-(340m/s)t_{rock}+ (3.6s)(340m/s) =0$

$(-4.9)\Delta t_{rock}^2-(340)t_{rock}+ 1224 =0$

The two possible values are and . Time cannot be a negative value so the time for the rock to fall is .

Using this information, and returning to the original kinematic equation it is possible to find the height of the cliff.

$\Delta x = \frac{1}{2}(-9.8m/s^2)(3.43s)^2$

$\Delta x= -57.66m$

The negative indicates that the stone fell from its original position. Therefore the height of the cliff is

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Question

Leslie rolls a ball out of a window from above the ground, such that the initial y-velocity is zero. How long will it be before the ball hits the ground?

Answer

We are given the initial velocity, acceleration, and distance traveled. Using the equation below, we can solve for the time. Remember that the initial velocity is 0m/s so it drops out of the equation.

$\Delta x = v_0 \Delta t+\frac{1}{2}a\Delta t^2$

$\Delta x =\frac{1}{2}a\Delta t^2$

$\Delta t = \sqrt{\frac{2\Delta x}{a}}$

$\Delta t = \sqrt{\frac{2(0m-10m)}{(-9.8m/s^2)}}$

The distance is negative, which makes since because the ball is traveling downward. Also when taking the square root, only the positive value is needed as it is impossible to have negative time.

$\Delta t = 1.43s$

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Question

An object is dropped from the roof of a building, how fast is it traveling after ? How far would it have fallen? Assume the building is tall enough for the object to have not hit the ground during this time and neglect air resistance. Assume the acceleration due to gravity to be

Answer

The only force accelerating the object is gravity since it was dropped, not thrown. Thus, to find out the speed of the object after some time, simply multiply the time the object has fallen by the acceleration of gravity. We will use . Then use the average velocity to calculate the distance the phone fell.

Final velocity after :

Distance the phone has fallen during the 9s of free fall:

$\Delta x = v_0 \Delta t+\frac{1}{2}a\Delta t^2$

Remember that the initial velocity of the phone is 0m/s. This can be removed from the equation.

$\Delta x =\frac{1}{2}a\Delta t^2$

$\Delta x =\frac{1}{2}(10m/s^2)(9s)^2$

$\Delta x = 405m$

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Question

If you threw a tomato upwards with an initial velocity of , at what time (in seconds) would the tomato hit the ground? Assume the acceleration due to gravity is

Answer

Gravity accelerates everything downward by 10m/s2. When the tomato is thrown upward with some velocity, gravity immediately begins to slowly reduce this velocity since the acceleration opposes the direction of the velocity.

At the top of the peak, the velocity of the tomato is .

$v_f=v_0 +a\Delta t$

Rearrange for time

$\Delta t = \frac{v_f-v_0}a{}$

$\Delta t = \frac{0m/s-12m/s}{-10m/s^2}$

$\Delta t = 1.2s$

In the tomato will reach its maximum height. If you have ever thrown a ball upward you may have noticed how it appears to stop at the peak. We have just calculated the time it takes for that ball to appear to stop for a very small time and fall back down. Since our tomato must travel back to Earth, we double the time for its up and down motion (since they equal each other) to get as the final answer.

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Question

A ball is thrown vertically with a velocity of . What is its velocity at the highest point in the throw?

Answer

When examining vertical motion, the vertical velocity will always be zero at the highest point. At this point, the acceleration from gravity is working to change the motion of the ball from positive (upward) to negative (downward). This change is represented by the x-axis on a velocity versus time graph. As the ball changes direction, its velocity crosses the x-axis, momentarily becoming zero.

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Question

Two balls, one with mass and one with mass , are dropped from above the ground. Which ball hits the ground first?

Answer

The mass of an object is completely unrelated to its free-fall motion. The equation for the vertical motion for an object in freefall is:

$\Delta x = v_0 \Delta t+\frac{1}{2}a\Delta t^2$

Notice, there is no mention of mass anywhere in this equation. The only thing that affects the time an object takes to hit the ground is the acceleration due to gravity and the distance travelled. Since these objects travel the same distance and are affected by the same gravitational force, they will fall for the same amount of time and hit the ground together.

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Question

A tennis ball is thrown straight up and it is caught at the same height the person released the ball from their hand. Which of the following is false? Ignore air resistance.

Answer

The question asks to point out the false statement. Everything on Earth is accelerated downwards by gravity, all the time, by . Think of gravity as having a negative sign. When the ball is thrown up, acceleration is working against the velocity slowing the ball down. Their signs are opposite. But when the ball is falling back down to Earth, the velocity and acceleration have the same sign. So velocity and acceleration will not always have the same sign.

If gravity could be said to have a negative sign since it pulls everything downward, then an upward velocity would have a positive (and opposing sign). At the top of the trajectory when the ball's upward velocity is finally overcome by gravity, the sign of the velocity becomes negative as it now points back down to Earth. So it is true velocity changes sign at the top.

Since the ball is caught at the same height, both the time up and time down are equal and it will be traveling at the same speed. Since gravity is the only force acting on it, the ball loses all its velocity on the way up and regains that exact amount by the time it reaches the height it started the journey. This also makes the time up equal to the time it falls. If the same force acts with same strength (gravity) the entire time, why would either of these change?

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Question

A runner wants to complete a run in less than . After running at constant speed for exactly , the runner still has left to run. The runner must then accelerate at for how many seconds in order to reach a final velocity that will allow them to complete the left of the race in the desired time?

Answer

Knowns:

$\Delta x_{total} = 10,000m$

$\Delta t_{total} = 30 min$

$\Delta t_{constant speed} = 26 min$

$\Delta x_{accelerating} = 1500m$

Unknown:

$\Delta t_{accelerating}= ?$

Equation:

The first thing is to determine the initial velocity of the runner before the runner accelerates for the final portion of the race. Since the runner is traveling at a constant velocity

$v = \frac{\Delta x}{\Delta t}$

Next, convert the time during the constant speed portion to seconds.

$26 min \frac{60s}{1 min}= 1560s$

Determine the amount of distance traveled while at a constant speed.

Use these values to determine the velocity of the runner.

$v = \frac{8,500m}{1560s}$

Next determine the final velocity needed for the runner to finish out the race in the remaining time.

left in race

$4min \frac{60s}{1 min} = 240s$

$v = \frac{1500m}{240s}$

Finally use the kinematic equations to calculate the time needed to get to this velocity.

$v_f= v_0+a\Delta t$

Rearrange for time

$\Delta t = \frac{v_f-v_0}{a}$

$\Delta t = \frac{6.25m/s-5.45m/s}{0.30m/s^2}$

$\Delta t = 2.67s$

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Question

Trevor is traveling when he sees a red light ahead. His car is capable of decelerating at a rate of . If it takes him to get the brakes on and he is from the intersection when he sees the light, how far from the beginning of the intersection will he be, and in what direction? In other words, will he be able to stop in time?

Answer

Knowns:

$t_{reaction} = 0.360s$

$\Delta x = 19m$

Unknowns:

$\Delta x$ that Trevor actually travels = ?

Equation:

Since Trevor takes a moment to react before stepping on the breaks, determine how far Trevor travels during the reaction time. At this time he is traveling at a constant velocity.

$v = \frac{\Delta x}{\Delta t}$

$\Delta x = v\Delta t$

$\Delta x = 20m/s 0.360s$

$\Delta x = 7.2m$

Next, using kinematic equations to determine where Trevor stops his car based on the acceleration of his car.

$v^f_2=v^0_2 +2a\Delta x$

$\Delta x =\frac{v^f_2-v^0_2}{2a}$

$\Delta x =\frac{0^2-(20m/s)^2}{2(-3.45m/s^2)}$

$\Delta x = 57.97m$ beyond the reaction point

Total distance traveled is the reaction time distance plus the distance traveled during the deceleration period.

total distance traveled

To find the distance beyond the red light, subtract the distance traveled from the distance to the light

distance beyond red light

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Question

Along a highway there is an unmarked police car traveling a constant . The officer is passed by a speeder traveling . Precisely after the speeder passes, the officer steps on the accelerator to catch the speeder. If the police car’s acceleration is , how much time passes before the police car overtakes the speeder? Assume here that the speeder is moving at a constant speed.

Answer

Knowns:

$v_{speeder} = 130km/h$

$v_{officer} =900km/h$

$t_{0 speeder} = 0 seconds$

$t_{0 officer}= 1 second$

Unknowns:

$t_{final}=?$

Equation:

To make things easier, set the frame of reference to the police officer. This means that the police officer is traveling at 0km/h and the speeder is moving at a speed relative to the officer.

$v_{speeder in reference to officer} = v_{speeder}- v_{officer}$

So the new relative velocities are:

$v_{speeder} = 40km/h$

$v_{officer} =0km/h$

This will help make the math easier.

Convert the relative velocity of the speeder to m/s.

$\frac{40km}{h}\frac{1000m}{1km}\frac{1h}{60 min}\frac{1min}{60s}= 11.11m/s$

Now there are two equations that have to be true for the officer to catch up to the speeder.

$\Delta x_{officer} =v_o\Delta t_{officer} + \frac{1}{2}a\Delta t^2_{officer}$

$v_{speeder}=\frac{\Delta x_{speeder}}{\Delta t_{speeder}}$

There are two things that connect these equations. First, the distance that the officer travels and the speeder travels must be the same. Additionally the final velocity of both the officer and the speeder must be the same.

Rearrange the speeder equation for the distance traveled.

$v_{speeder} \Delta t_{speeder} = \Delta x_{speeder}$

Set the two equations equal, as the distance traveled of the speeder is equal to the distance traveled by the officer.

$v_o\Delta t_{officer} + \frac{1}{2}a\Delta t^2_{officer}=v_{speeder} * \Delta t_{speeder}$

Remember that the reference point is at the officer’s perspective so the officer has an initial velocity of 0m/s.

$0m/s + \frac{1}{2}a\Delta t^2_{officer}=v_{speeder} * \Delta t_{speeder}$

Rewrite the equation breaking up the change of time into time final and time initial.

$\frac{1}{2}a(t_{f officer}- t_0 officer)^2=v_{speeder} * (t_{f speeder} -t_{0 speeder})$

Distribute on the left side.

$\frac{1}{2}a(t^2_{f officer}- t_{0 officer}t_{f officer} +t^2_{0 officer})=v_{speeder} * (t_{f speeder} -t_{0 speeder})$

$\frac{1}{2}at^2_{f officer}- \frac{1}{2}at_{0 officer}t_{f officer} +\frac{1}{2}at^2_{0 officer}=v_{speeder} * (t_{f speeder} -t_{0 speeder})$

Substitute values

$\frac{1}{2}(2.5m/s^2)t^2_{f officer}- \frac{1}{2}(2.5m/s^2)(1s)t_{f officer} +\frac{1}{2}(2.5m/s^2) (1s)^2=11.11m/s* (t_{f speeder} -0s)$

Simplify

$(1.25)t^2_{f officer}- 1.25t_{f officer} +1.25=11.11 t_{f speeder}$

Remember that the final time for both the officer and the speeder are the same. So rearrange this in the form of a quadratic equation.

$(1.25)t^2_{f officer}- 12.36t_{f officer} +1.25=0$

Use the quadratic formula to solve.

Possible answers: or

Since the officer did not move until , then the second answer must be correct.

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Question

A ball rolls to a stop after . If it had a starting velocity of , what is the deceleration on the ball due to friction?

Answer

Explanation:

We are given the initial velocity, time, and final velocity (zero because the ball stops). Using these values and the appropriate motion equation, we can solve for the acceleration.

Acceleration is given by the change in velocity over time:

$a = \frac{v_f-v_0}{\Delta t}$

We can use our values to solve for the acceleration.

$a = \frac{0m/s - 8.1m/s}{4.2s}$

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Question

Usain Bolt accelerates from rest to over a distance of . What is his acceleration? Assume the acceleration is constant.

Answer

We are given final velocity and we know initial velocity is 0. We can use the following kinematics formula to relate final speed, initial speed, acceleration, and displacement:

$v_f^2=v_0^2+2a\Delta x$

The initial velocity is zero since Usain starts from rest. Therefore we can remove it from the equation.

$v_f^2=2a\Delta x$

Rearrange the equation to solve for acceleration.

$\frac{v_f^2}{2\Delta x}=a$

$\frac{(13.4m/s)^2}{2(100m)}=a$

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Question

Peter starts from rest and runs down a hallway . If his final velocity is , how far did he run?

Answer

Knowns:

$\Delta t =31s$

Unknowns:

$\Delta x = ?$

Equation:

First we need to determine Peter’s acceleration. The best kinematic equation to use here is:

$v_f=v_0+a\Delta t$

$\frac{v_f-v_0}{\Delta t}=a$

$\frac{13m/s-0m/s}{31s}=a$

Once Peter’s acceleration is known it is possible to find the distance he traveled.

$\Delta x = v_0 \Delta t+\frac{1}{2}a\Delta t^2$

Peter’s initial velocity is so this equation simplifies to

$\Delta x =\frac{1}{2}a\Delta t^2$

$\Delta x =\frac{1}{2}(0.42m/s^2)(31)^2$

$\Delta x = 201.8m$

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Question

Objects and both start at rest. They both accelerate at the same rate. However, object accelerates for twice the time as object . What is the final speed of object compared to that of object ?

Answer

$v_f = v_0+a\Delta t$

There is a direct relationship between the final velocity and the time traveled. Therefore if the time is doubled, the velocity would double as well.

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Question

Objects and both start from rest. They both accelerate at the same rate. However, object accelerates for twice the time as object . What is the distance traveled by object compare to that of object ?

Answer

$\Delta x = v_0 \Delta t+\frac{1}{2}a\Delta t^2$

The distance traveled is directly related to the square of the time traveled. Therefore time if time is doubled, the value will be squared and therefore four times as great. The distance traveled will be 4 times as far.

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Question

Suppose a can is kicked and then travels up a smooth hill of ice. Which of the following is true about its acceleration?

Answer

The can will have the same acceleration both up and down the hill since there is no friction. Friction or other forces would cause a change in acceleration. But since there is no friction, the can will travel up the hill, slow down and then accelerate back down the hill.

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Question

Leslie rolls a ball out of a window from 10 meters above the ground, such that the initial y-velocity is zero. How long will it be before the ball hits the ground?

$\small g=-9.8\frac{m}{s^2}$

Answer

We are given the initial velocity, acceleration, and distance traveled. Using the equation below, we can solve for the time.

$\Delta y=V_i+\frac{1}{2}at^2$

$(y_f-y_i)=V_i+\frac{1}{2}at^2$

$(0m-10m)=0\frac{m}{s}+\frac{1}{2}(-9.8\frac{m}{s^2})t^2$

$(-10m)=(-4.9\frac{m}{s^2})t^2$

The distance is negative, which makes since because the ball is traveling downward.

$\frac{-10m}{-4.9\frac{m}{s^2}}=t^2$

$2.04s^2\approx t^2$

$\sqrt{2.04s^2}\approx \sqrt{t^2}$

$1.43s\approx t$

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Question

Derek rolls a ball along a flat surface with an initial velocity of $\small 2.2\frac{m}{s}$ . If it stops after 12 seconds, what was the acceleration on the ball?

Answer

Since the ball starts with a positive velocity and ends at rest, we can predict that the acceleration will be negative. Using the values given in the question and the equation below, we can solve for the acceleration.

$0\frac{m}{s}=2.2\frac{m}{s}+a(12s)$

$\frac{-2.2\frac{m}{s}}{12s}=a$

$-0.183\frac{m}{s^2} \approx a$

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