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AP Physics B : Linear Motion

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65 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

Example Question #1 : Motion In One Dimension

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

Possible Answers:

125m

250m

75m

50m

100m

Correct answer:

125m

Explanation:

Using the equation d=v_ot + (1/2)a(t^2) 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, v_o =0 , thus we have d=(1/2)a(t^2) . When we plug in our values, and assuming that acceleration is equal to gravity (10m/s²) we find that d=(1/2)(10)(5^2) = 125m.

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Example Question #1 : Motion In One Dimension

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

Possible Answers:

50m

250m

300m

500m

25m

Correct answer:

500m

Explanation:

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 d=v_ot +(1/2)at^2 can be used to find the distance traveled.

d=0+(1/2)(10m/s^2)(10s)^2=500m

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Example Question #1 : Motion In One Dimension

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

Possible Answers:

7.3s

12.6s

10.9s

5.5s

2.7s

Correct answer:

5.5s

Explanation:

Using the equation d=v_ot + (1/2)at^2 , we can solve for time.

Since the object started at rest, v_0=0 . Now we are left with the equation d=(1/2)at^2 .

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

30s^2=t^2

t=5.5s

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

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Example Question #1 : Motion In One Dimension

A person on top of a 200m tall building drops a rock. How long will it take for the rock to reach the ground? Ignore air resistance.

Possible Answers:

t=113.4s

t=7.3s

t=40.8s

t=55.7s

t=6.4s

Correct answer:

t=6.4s

Explanation:

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 y=0 and $y_o=200\ \text{m}$ .

Also, the initial speed is zero.

v_o=0

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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Example Question #3 : Motion In One Dimension

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 28m 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?

Possible Answers:

$a=-5.12\ \frac{\text{m}}{\text{s}^2}$

$a=0\ \frac{\text{m}}{\text{s}^2}$

$a=-4.51\ \frac{\text{m}}{\text{s}^2}$

$a=-13.97\ \frac{\text{m}}{\text{s}^2}$

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

Correct answer:

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

Explanation:

Use the following kinematic equation where the initial velocity is v_o , final velocity is v_f , 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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Example Question #1 : Linear Motion And Momentum

A 2.5kg is dropped from a height of 20m . A picture is taken when the ball is from the ground with an exposure time of 0.005s . If the actual diameter of the ball is 15cm , what will the vertical diameter of the ball appear to be in the picture?

Possible Answers:

23.6cm

36.8cm

19.9cm

15cm

17.2cm

Correct answer:

23.6cm

Explanation:

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 0.086m , or 8.6cm . The diameter of the ball in the vertical direction will appear to be distorted by this distance.

15cm+8.6cm=23.6cm

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Example Question #3 : Motion In One Dimension

A person stands on the edge of a straight 20 m -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, 20 m below?

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

Possible Answers:

43.00 s

16.12 s

5.334 s

2.562 s

6.433 s

Correct answer:

2.562 s

Explanation:

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 20 m below where is started, so (x - x_o) = -20 m .

The initial velocity, v_o , 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 -1.562 s ).

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

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Example Question #311 : Ap Physics 1

A block is placed at a height of 18m on a ramp at an incline of 24^o . 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?

24_deg._incline_block

Possible Answers:

$v_f=5.1\frac{m}{s}$

$v_f=16\frac{m}{s}$

$v_f=8.9\frac{m}{s}$

$v_f=14\frac{m}{s}$

$v_f=12\frac{m}{s}$

Correct answer:

$v_f=14\frac{m}{s}$

Explanation:

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 v_o=0 . 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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Example Question #281 : Mcat Physical Sciences

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?

Possible Answers:

70m

280m

140m

420m

Correct answer:

140m

Explanation:

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, v_y, 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 v_f = v_o +at .

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

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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Example Question #2 : Motion In Two Dimensions

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

Possible Answers:

675m

1350m

62.5m

250m

Correct answer:

1350m

Explanation:

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.

v_f = v_o + at

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

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 = v_x(t)

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

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AP Physics B : Linear Motion

Study concepts, example questions & explanations for AP Physics B

All AP Physics B Resources

Example Questions

Example Question #1 : Motion In One Dimension

Example Question #1 : Motion In One Dimension

Example Question #1 : Motion In One Dimension

Example Question #1 : Motion In One Dimension

Example Question #3 : Motion In One Dimension

Example Question #1 : Linear Motion And Momentum

Example Question #3 : Motion In One Dimension

Example Question #311 : Ap Physics 1

Example Question #281 : Mcat Physical Sciences

Example Question #2 : Motion In Two Dimensions

All AP Physics B Resources

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