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AP Physics 1 : AP Physics 1

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Example Questions

Example Question #42 : Motion In One Dimension

A basketball player jumps directly upward with an initial velocity of $18\frac{m}{s}$ . Neglecting air resistance, how long does he spend in the air?

Possible Answers:

4.4 s

4.0 s

3.6 s

4.2 s

3.8 s

Correct answer:

3.6 s

Explanation:

Use a kinematic equation to solve for the time.

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

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

$0\frac{m}{s}=18\frac{m}{s}+(-9.8\frac{m}{s^{2}})(t_{f}-0s)$

$-18\frac{m}{s}= (-9.8\frac{m}{s^{2}})(t_{f})$

$t_{f}=1.8 s$

Since that describes half of the time he spends in the air you must multiply by 2 for the final answer.

$t_{f}=3.6 s$

The basketball player will be int he air for a total of 3.6 s .

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Example Question #46 : Linear Motion And Momentum

A rubber ball bounces off the floor at $45\frac{m}{s}$ , how high will the ball be at it's apex? For this question neglect air resistance.

Possible Answers:

103 m

112 m

106 m

110 m

100 m

Correct answer:

110 m

Explanation:

Use a kinematic equation to solve for the final time.

$V_f=Vo+a(\Delta t)$

Plug in the known values and solve for time final.

$0\frac{m}{s}=45\frac{m}{s}+(-9.8\frac{m}{s^{2}})(\Delta t)$

$-45\frac{m}{s}=(-9.8\frac{m}{s^{2}})(t_{f})$

$t_{f}=4.59 s$

Use another kinematic equation and solve for the final distance.

$X_f=X_i+V_o(\Delta t)+0.5(a)(\Delta t)^{2}$

Plug in and solve for the final distance

$X_{f}=0m+45\frac{m}{s}(4.59s)+\frac{1}{2}(-9.8\frac{m}{s^{2}})(4.59s)^{2}$

$X_{f}=206.55m+(-4.9\frac{m}{s^{2}})(21.07s^{2})$

$X_{f}=206.55m+(-96.7m)$

$X_{f}=109.9m$

The rubber ball will be 110 m off the ground at it's highest.

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Example Question #47 : Linear Motion And Momentum

A train starting in Denver, proceeds down a straight track from rest at a constant acceleration of $1.25\frac{m}{s^2}$ for exactly 1 min . The train then goes at that speed for another 5 min . Neglecting both the effects of friction and air resistance how far did the train get in kilometers away from Denver?

Possible Answers:

24.8 km

26.0 km

25.2 km

23.6 km

24.2 km

Correct answer:

24.8 km

Explanation:

Use the equation for acceleration to solve for the final velocity.

$a=\frac{(V_f-V_i)}{(t_f-t_i)}$

Plug in and solve for the final velocity.

$1.25\frac{m}{s^{s}}=\frac{V_{f}-0\frac{m}{s}}{60s-0s}$

$V_{f}=75\frac{m}{s}$

use another kinematic equation to solve for the final distance

$X_f=X_i+V_o(\Delta t)+0.5(a)(\Delta t)^{2}$

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

$X_{f}=0m+(75\frac{m}{s})(300s)+\frac{1}{2}(1.25\frac{m}{s^{2}})(60s)^{2}$

$X_{f}=22500m+2250m$

$X_{f}=24750m$

The answer needs to be in kilometers so divide by 1000m

$\frac{24750m}{1000m}=24.75 km$

Using significant figures 24.8km .

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Example Question #41 : Linear Motion And Momentum

My roommate locked himself out of the apartment and forgot his keys. He asks me to walk over to the balcony and drop my key off the edge so he can get in. The keys take 6.0 s to reach the ground from the time it is released from my hand. From what height were the keys released, neglect the effects due to air resistance?

Possible Answers:

180 m

172 m

176 m

178 m

174 m

Correct answer:

176 m

Explanation:

Use a kinematic equation to solve for the height of the building.

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

Plug in and solve for the height of the building.

$0m=X_i+0\frac{m}{s}(6.0s)+\frac{1}{2}(-9.8\frac{m}{s^{2}}(6.0s)^{2}$

$0m=X_i+(-4.9\frac{m}{s^{2}})(36s^{2})$

X_i= 176.4m

The building was 176 m tall.

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Example Question #49 : Linear Motion And Momentum

A leaf falls from an Aspen tree at an initial height of 20 m . The leaf maintains a high surface to volume ratio and we can not neglect air resistance. Now, the new free fall acceleration due to gravity is equal to 75% it's usual value. How long does it take the leaf to fall directly to the ground bellow?

Possible Answers:

5 s

1 s

2 s

4 s

3 s

Correct answer:

2 s

Explanation:

Solve for the new acceleration due to gravity.

$a=(-9.80\frac{m}{s^{2}})(0.75)$

$a=-7.35\frac{m}{s^{2}}$

Use a kinematic equation to solve for the final time.

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

Plug in and solve for the final time.

$0m=20m+0\frac{m}{s}(\Delta t)+\frac{1}{2}(-7.35\frac{m}{s^{2}})(\Delta t)^{2}$

$-20m=-3.675\frac{m}{s^{2}}(t_{f}-0s)^{2}$

$5.44s^{2}=t_{f}^{2}$

$t_{f}=2.33 s$

Using significant figures and rounding the answer becomes 2 s .

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Example Question #41 : Motion In One Dimension

A rocket ship is launched straight up from rest with a constant acceleration of $36\frac{m}{s^2}$ . After 200 seconds, how fast is the rocket ship traveling? Ignore both the effects of friction and of acceleration due to gravity.

Possible Answers:

$9600\frac{m}{s}$

$2700\frac{m}{s}$

$1100\frac{m}{s}$

$4800\frac{m}{s}$

$7200 \frac{m}{s}$

Correct answer:

$7200 \frac{m}{s}$

Explanation:

Use the following kinematic formula to find the final velocity after 200 seconds:

V_f= V_o + at

Plug in known values and solve.

$V_f=0\frac{m}{s}+36\frac{m}{s^2}*200s$

$V_f=7200\frac{m}{s}$

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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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AP Physics 1 : AP Physics 1

Study concepts, example questions & explanations for AP Physics 1

All AP Physics 1 Resources

Example Questions

Example Question #42 : Motion In One Dimension

Example Question #46 : Linear Motion And Momentum

Example Question #47 : Linear Motion And Momentum

Example Question #41 : Linear Motion And Momentum

Example Question #49 : Linear Motion And Momentum

Example Question #41 : Motion In One Dimension

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

All AP Physics 1 Resources

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