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AP Physics 1 : Linear Motion and Momentum

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Example Questions

Example Question #51 : Linear Motion And Momentum

A car is moving in the eastward direction at $25 \frac{m}{s}$ when a skateboard rolls into the road. The driver is forced to apply his brakes causing an acceleration of $-2.3 \frac{m}{s^{2}}$ . How long does it take the car to come to a complete stop?

Possible Answers:

9 s

12 s

13 s

11 s

10 s

Correct answer:

11 s

Explanation:

Use one of the kinematic equations.

$V_f=V_+a\Delta t$

Plug in the known values and solve.

$0\frac{m}{s}=25\frac{m}{s}+(-2.3\frac{m}{s^{2}})(\Delta t)$

$-25\frac{m}{s}=-2.3\frac{m}{s^{2}}(t_{f}-0s)$

t_f= 11seconds

The car will take 11 seconds to come to a complete stop.

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Example Question #52 : Linear Motion And Momentum

A wreak-less card driver was driving in the eastward direction at $65\frac{m}{s}$ when he noticed that the car in front of him was at a complete halt. He subsequently slammed the brakes causing an acceleration of $-7.5\frac{m}{s^{2}}$ . Will he be saved by his brakes or will he hit the car that was 500 m in front of him when he first applied the brakes?

Possible Answers:

Hits the car; or would stop at 550 m

Stops just safely at 490 m

Stops safely at 450 m

Hits the car; or would stop at 600 m

Would hit the car; or stop at 530 m

Correct answer:

Would hit the car; or stop at 530 m

Explanation:

Use a kinematic equation to solve for the total time.

$V_f=V_o+a\Delta t$

Plug in and solve for time.

$0\frac{m}{s}=65\frac{m}{s}+(-7.5\frac{m}{s^{2}})(t_{f}-0s)$

$-65\frac{m}{s}=(-7.5\frac{m}{s^{2}})(t_{f})$

$t_{f}=8.66 s$

Use another kinematic equation to solve for the final distance.

$X_f=X_i+V_o(\Delta t)+\frac{1}{2}(a)(\Delta t)^{^{2}}$

Plug in the known values and solve for the final distance.

$X_{f}=0m+65\tfrac{m}{s}(8.66s)+\frac{1}{2}(-7.5\frac{m}{s}^{2})(8.66s)^{2}$

$X_{f}=563.3m+(-32.47m)$

$X_{f}=530.8 m$

The car would have stopped at 530.8 m , however the other car was at 500 m thus there would have unfortunately been a car accident.

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Example Question #51 : Motion In One Dimension

Two cars leave Colorado Springs at 3:00 pm. Car "A" travels at $25 \frac{m}{s}$ for 2 h then accelerates at $1.5\frac{m}{s^{2}}$ for 15 s . Then continues on at that speed until it arrives in Arizona about 1000 km away. If the driver of car "B" would like to leave from Colorado Springs at the same time as car "A" at what average velocity would he/she have to drive to arrive at the same time as car "A"? Ignore both the effects of air resistance and friction.

Possible Answers:

$37\frac{m}{s}$

$41\frac{m}{s}$

$40\frac{m}{s}$

$39\frac{m}{s}$

$38\frac{m}{s}$

Correct answer:

$41\frac{m}{s}$

Explanation:

Use the equation for velocity to determine the final distance of car "A" in it's initial movement.

$V=\frac{(X_f-X_i)}{(t_f-t_i)}$

Plug in the known values and solve for the final distance.

$25\frac{m}{s}=\frac{(X_{f}-0m)}{(7200s-0s)}$

$X_{f}=180000m$

Then solve for the final velocity after the short period of acceleration.

$V_f=V_o=a(\Delta t)$

$V_{f}=25\frac{m}{s}+1.5\frac{m}{s^{2}}(15s)$

$V_{f}=47.5\frac{m}{s}$

Now, solve for the total distance car "A" traveled before and after the period of acceleration.

$X_f=X_i+V_o(\Delta t)+\frac{1}{2}(a)(\Delta t)^{2}$

Plug in and solve for the final distance.

$X_{f}=0m+25\frac{m}{s}(7200s)+\frac{1}{2}(1.5\frac{m}{s^{2}})(15s)^{2}$

$X_{f}=180000m+168.75m$

$X_{f}=180168.75m$

The entire trip is 1000km or 1000000m

Distance Remaining=1000000m-180168.75m

DistanceRemaining=819831.25m

$V=\frac{\Delta X}{\Delta t}$

$\Delta t = \frac{\Delta X}{ V}$

$\frac{819831.25m}{47.5\frac{m}{s}}$

t=17259.6s

t(total)= 24474.6s

Now use the velocity equation to solve for the average velocity car "B" must travel.

$V=\frac{\Delta X}{\Delta t}$

$V=\frac{1000000m-0m}{24474.6s-0s}$

$V=40.86\frac{m}{s}$

Using significant figures the answer is $41\frac{m}{s}$ .

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Example Question #54 : Linear Motion And Momentum

A bicyclist is traveling down a straight street at a velocity of $4\frac{m}{s}$ . At time t=0 , the bicyclist reaches an incline. Despite pedaling as hard as he can, the bicyclist begins decelerating at a rate of $a = 0.2\frac{m}{s^2}$ . How far has the bicyclist traveled when t = 15s ?

Possible Answers:

60m

30m

82.5m

22.5m

37.5m

Correct answer:

37.5m

Explanation:

We can use a kinematics equation to solve this problem:

$x_f = x_i + v_it+\frac{1}{2}at^2$

Rearranging for change in distance:

$\Delta x = v_it+\frac{1}{2}at^2$

We have all of the values we need, so we can solve the problem:

$\Delta x = \left ( 4\frac{m}{s}\right )(15s)+\frac{1}{2}\left ( -0.2\frac{m}{s^2}\right )(15s)^2$

Note that the acceleration is negative since she is decelerating from a positive velocity.

$\Delta x = 37.5m$

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Example Question #52 : Motion In One Dimension

Consider the following scenario:

Sledder

A sledder of mass is at the stop of a sledding hill at height with a slope of angle .

A sledder is accelerating down the hill that has a slope $20^{\circ}$ at a rate of $4\frac{m}{s^2}$ . If the sledder has an initial velocity of $2\frac{m}{s}$ , how far has the sledder traveled along the hill if her final velocity is $8\frac{m}{s}$ ?

Possible Answers:

7.5m

16m

12m

32m

Correct answer:

7.5m

Explanation:

We can use the follow kinematics equation to solve this problem:

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

Rearranging for change in x, we get:

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

Now plugging in values for each variable:

$\Delta x = \frac{\left ( 8\frac{m}{s}\right )^2-\left ( 2\frac{m}{s}\right )^2}{2\left ( 4\frac{m}{s^2}\right )}$

$\Delta x = 7.5m$

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Example Question #53 : Motion In One Dimension

Consider the following scenario:

Sledder

A sledder of mass is at the stop of a sledding hill at height with a slope of angle .

If the sledder is traveling down a hill which has a slope $\angle 40^{\circ}$ and has a constant horizontal velocity of $v_x = 6\frac{m}{s}$ , how much time does it take for the sledder to drop a vertical distance of 12m .

Possible Answers:

0.3s

3.6s

1.1s

5.8s

2.4s

Correct answer:

2.4s

Explanation:

To solve this problem we will go in the following steps:

1. Calculate linear velocity parallel with slope

2. Calculate vertical component of velocity

3. Calculate time it takes to drop 12m

So first, we will use the cosine function:

$cos(\theta)=\frac{v_x}{v}$

Rearranging:

$v = \frac{v_x}{cos(\theta)}$

Plugging in our value:

$v = \frac{\left ( 6\frac{m}{s}\right )}{cos(40^{\circ})} = 7.83\frac{m}{s}$

Now to calculate the vertical component, we will use the sine function:

$sin(\theta)=\frac{v_y}{v}$

Rearranging

$v_y = vsin(\theta)$

$v_y = (7.83)sin(40^{\circ}) = 5.03\frac{m}{s}$

Now using this to calculate the time it takes to drop 12m :

$\Delta y = v_yt$

$t = \frac{\Delta y}{v_y} = \frac{12m}{5.03\frac{m}{s}}$

t = 2.4s

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Example Question #57 : Linear Motion And Momentum

An arrow is shot straight up in the air with an initial velocity of $35\frac{m}{s}$ from a height of. How long does it take for the arrow to hit the ground? Neglect air resistance.

$g = 10\frac{m}{s^2}$

Possible Answers:

7.1s

5.8s

6.4s

7.0s

2.0s

Correct answer:

7.1s

Explanation:

Since the arrow is shot straight up in the air, we can calculate how long it takes for the arrow to reach its high point (where it will have a velocity of 0):

v_f = v_i +at

$0\frac{m}{s}= 35\frac{m}{s}+\left ( -10\frac{m}{s^2}\right )t$

t = 3.5 s

Furthermore, this will be the same amount of time it takes to get to its original position. Then we just need to calculate how long it takes the arrow to travel from its original position to the ground, using the follow kinematics equation:

$\Delta y = v_{y_{i}}t+\frac{1}{2}at^2$

Plugging in our values:

$-2m = \left ( -35\frac{m}{s}\right )t+\frac{1}{2}\left ( -10\frac{m}{s^2}\right )t^2$

Rearranging:

$\left ( 5\frac{m}{s^2}\right )t^2+\left ( 35\frac{m}{s}\right )t-2m=0$

Using the quadratic equation, we get:

t = 0.057s or -7.06s

Since we can't have a negative time, the first option is the correct one. Adding this to our other time, we get:

t = 7.1s

Alternatively, we could have solved the equation by putting the initial conditions in the kinematic equation.

$-2m = \left ( 35\frac{m}{s}\right )t+\frac{1}{2}\left ( -10\frac{m}{s^2}\right )t^2$

However, the first method emphasizes the importance of being able to solve a problem multiple different ways and incorporates multiple concepts surrounding projectile motion.

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Example Question #51 : Linear Motion And Momentum

A physicist over-pressurized a water heater that exploded and shot into the air. The water heater stayed in the air for $12 \textup{ s}$ before it came crashing back to earth. How far above the ground did the water heater reach at it's tallest point? (Ignore air resistance.)

Possible Answers:

$60\textup{ m}$

$720\textup{ m}$

$360\textup{ m}$

$120\textup{ m}$

$180\textup{ m}$

Correct answer:

$180\textup{ m}$

Explanation:

We can solve these problems using kinematic equations. First we start with:

$v = v_{0} + at$

$0= v_{0} + (-10\frac{m}{s^{2}})(6s))$

$v_{0} = 60\frac{m}{s}$

We know that the final velocity is $0\frac{m}{s}$ because at the top of an arc the water heater doesn't have vertical movement. The acceleration is due to gravity, and the time is only half the time that was given in the problem (half the time is the water heater going up, the other half is it falling back to the ground).

When we know the initial velocity of the water heater we can then use:

$y = v_{0}t + \frac{1}{2}at^{2}$

$y=(60\frac{m}{s})(6s)+\frac{1}{2}(-10\frac{m}{s})(6s)^{2}$

y = 180m

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Example Question #59 : Linear Motion And Momentum

An object on frictionless wheels is being pulled with a force of 20N and accelerates at rate of $5 \frac{m}{s^{2}}$ . If a second object is also pulled at 20N but weighs twice as much as the first object, what would it's acceleration be?

Possible Answers:

$2.1 \frac{m}{s^{2}}$

$4.1 \frac{m}{s^{2}}$

$2.5 \frac{m}{s^{2}}$

$4.5 \frac{m}{s^{2}}$

Correct answer:

$2.5 \frac{m}{s^{2}}$

Explanation:

First we must find the original object's mass using the equation F=ma . . We then double the mass for the second object and again use . Solve for acceleration using Mass2 = 8kg and Force2 = 20N .

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Example Question #51 : Linear Motion And Momentum

A ball is dropped from a building and at one instant its speed is $25\frac{m}{s}$ . What is its approximate speed one second later?

Possible Answers:

$30\frac{m}{s}$

$35\frac{m}{s}$

$25\frac{m}{s}$

$40\frac{m}{s}$

Correct answer:

$35\frac{m}{s}$

Explanation:

The acceleration of an object due to gravity is $9.8\frac{m}{s^{2}}$ . If you multiply this value by one second, you can see that an object increases by a speed of $9.8\frac{m}{s}$ per second. After one second the ball is traveling $9.8\frac{m}{s}$ faster.

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AP Physics 1 : Linear Motion and Momentum

Study concepts, example questions & explanations for AP Physics 1

All AP Physics 1 Resources

Example Questions

Example Question #51 : Linear Motion And Momentum

Example Question #52 : Linear Motion And Momentum

Example Question #51 : Motion In One Dimension

Example Question #54 : Linear Motion And Momentum

Example Question #52 : Motion In One Dimension

Example Question #53 : Motion In One Dimension

Example Question #57 : Linear Motion And Momentum

Example Question #51 : Linear Motion And Momentum

Example Question #59 : Linear Motion And Momentum

Example Question #51 : Linear Motion And Momentum

All AP Physics 1 Resources

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