Call Now to Set Up Tutoring:

(888) 888-0446

AP Physics B : Motion

Study concepts, example questions & explanations for AP Physics B

Create An Account

All AP Physics B Resources

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:

75m

50m

250m

125m

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.

Report an Error

Example Question #1 : Linear Motion And Momentum

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

Possible Answers:

50m

25m

300m

500m

250m

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

Report an Error

Example Question #1 : Linear Motion And Momentum

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

Possible Answers:

5.5s

7.3s

10.9s

12.6s

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.

Report an Error

Example Question #11 : Newtonian Mechanics

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=6.4s

t=40.8s

t=7.3s

t=55.7s

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.

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

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

Report an Error

Example Question #2 : 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:

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

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

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

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

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

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.

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

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

Report an Error

Example Question #11 : Motion

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:

15cm

23.6cm

36.8cm

19.9cm

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

Report an Error

Example Question #1 : 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:

6.433 s

5.334 s

43.00 s

2.562 s

16.12 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.

Report an Error

Example Question #12 : Motion

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:

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

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

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

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

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

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.

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

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

Report an Error

Example Question #1 : Motion In Two Dimensions

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:

140m

70m

280m

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 $d = v_ot +\frac{1}{2}at^{2}$ .

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

Report an Error

Example Question #1 : 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:

1350m

250m

675m

62.5m

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)

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

Report an Error

View Tutors

Donald
Certified Tutor

Cleveland State University, Bachelor of Science, Chemistry. University of Southern California, Doctor of Philosophy, Organic ...

View Tutors

Victoria
Certified Tutor

Norte Dame de Namur University, Bachelor in Arts, Sociology. University of San Francisco, Master of Arts, Organizational Lead...

View Tutors

Shweta
Certified Tutor

All India Management Association , Master's/Graduate, Human Resource Management . Indian Law Institute , Master's/Graduate, C...

All AP Physics B Resources

65 Practice Tests Question of the Day Flashcards Learn by Concept

Popular Subjects

GMAT Tutors in San Diego, ACT Tutors in Atlanta, Algebra Tutors in Houston, French Tutors in New York City, Biology Tutors in Denver, Computer Science Tutors in Washington DC, SAT Tutors in Houston, Calculus Tutors in Houston, LSAT Tutors in Los Angeles, English Tutors in San Diego

Popular Courses & Classes

LSAT Courses & Classes in Chicago, LSAT Courses & Classes in Phoenix, GMAT Courses & Classes in Boston, GRE Courses & Classes in Washington DC, SAT Courses & Classes in Miami, GMAT Courses & Classes in Washington DC, ACT Courses & Classes in Denver, GRE Courses & Classes in San Francisco-Bay Area, SAT Courses & Classes in Atlanta, LSAT Courses & Classes in Atlanta

Popular Test Prep

GMAT Test Prep in Philadelphia, SSAT Test Prep in San Francisco-Bay Area, SAT Test Prep in Boston, SSAT Test Prep in Philadelphia, GRE Test Prep in Seattle, SAT Test Prep in Miami, LSAT Test Prep in Denver, ISEE Test Prep in Phoenix, GMAT Test Prep in Dallas Fort Worth, LSAT Test Prep in Houston

DMCA Complaint

If you believe that content available by means of the Website (as defined in our Terms of Service) infringes one or more of your copyrights, please notify us by providing a written notice (“Infringement Notice”) containing the information described below to the designated agent listed below. If Varsity Tutors takes action in response to an Infringement Notice, it will make a good faith attempt to contact the party that made such content available by means of the most recent email address, if any, provided by such party to Varsity Tutors.

Your Infringement Notice may be forwarded to the party that made the content available or to third parties such as ChillingEffects.org.

Please be advised that you will be liable for damages (including costs and attorneys’ fees) if you materially misrepresent that a product or activity is infringing your copyrights. Thus, if you are not sure content located on or linked-to by the Website infringes your copyright, you should consider first contacting an attorney.

Please follow these steps to file a notice:

You must include the following:

A physical or electronic signature of the copyright owner or a person authorized to act on their behalf; An identification of the copyright claimed to have been infringed; A description of the nature and exact location of the content that you claim to infringe your copyright, in \ sufficient detail to permit Varsity Tutors to find and positively identify that content; for example we require a link to the specific question (not just the name of the question) that contains the content and a description of which specific portion of the question – an image, a link, the text, etc – your complaint refers to; Your name, address, telephone number and email address; and A statement by you: (a) that you believe in good faith that the use of the content that you claim to infringe your copyright is not authorized by law, or by the copyright owner or such owner’s agent; (b) that all of the information contained in your Infringement Notice is accurate, and (c) under penalty of perjury, that you are either the copyright owner or a person authorized to act on their behalf.

Send your complaint to our designated agent at:

Charles Cohn Varsity Tutors LLC
101 S. Hanley Rd, Suite 300
St. Louis, MO 63105

Or fill out the form below:

Do not fill in this field

Contact Information

* Full legal name

Company name

* Phone number

* Email address

Address

* Address1

Address2

* City

* State

* Zipcode

* Country

Complaint Details

* URL of copyrighted work (where your original material is located)

* Please describe the copyrighted work so that it may be easily identified

I am the owner, or an agent authorized to act on behalf of the owner of an exclusive right that is allegedly infringed.

I have a good faith belief that the use of the material in the manner complained of is not authorized by the copyright owner, its agent, or the law.

This notification is accurate.

I acknowledge that there may be adverse legal consequences for making false or bad faith allegations of copyright infringement by using this process.

* Signature (your digital signature is legally binding)

HIGH SCHOOL

GRADUATE SCHOOL

K-8

Search 50+ Tests

math tutoring

science tutoring

foreign languages

elementary tutoring

other

Search 350+ Subjects

AP Physics B : 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 : Linear Motion And Momentum

Example Question #1 : Linear Motion And Momentum

Example Question #11 : Newtonian Mechanics

Example Question #2 : Motion In One Dimension

Example Question #11 : Motion

Example Question #1 : Motion In One Dimension

Example Question #12 : Motion

Example Question #1 : Motion In Two Dimensions

Example Question #1 : Motion In Two Dimensions

All AP Physics B Resources

Report an issue with this question

DMCA Complaint

Contact Information

Address

Complaint Details

Email address:
Your name:
Feedback: