Call Now to Set Up Tutoring:

(888) 888-0446

AP Physics 1 : Linear Motion and Momentum

Study concepts, example questions & explanations for AP Physics 1

Create An Account

All AP Physics 1 Resources

7 Diagnostic Tests 170 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

Example Question #31 : Linear Motion And Momentum

A car is traveling on the freeway at $36 \frac{m}{s}$ when the driver sees traffic stopped 100 m ahead. What acceleration is needed to stop the car safely?

Possible Answers:

$-6.5 \frac{m}{s^2}$

$13.0 \frac{m}{s^2}$

$0.18 \frac{m}{s^2}$

$-0.18 \frac{m}{s^2}$

$6.5 \frac{m}{s^2}$

Correct answer:

$-6.5 \frac{m}{s^2}$

Explanation:

We begin by noting we have been given no information about the time in which the stopping is to occur, only velocities and distance. This points us to the following kinematic equation:

$V_f^2=V_i^2+2a\Delta x$

By using the given information, and noting that coming to a stop implies a final velocity of $0 \frac{m}{s}$ , we can directly substitute the numbers in and solve. It makes sense the final value is negative because the car is being accelerated in the negative direction in order to stop it.

$a=\frac{-(36\frac{m}{s})^2}{2 (100m)}=-6.5 \frac{m}{s^2}$

Report an Error

Example Question #31 : Linear Motion And Momentum

An airplane is flying at $37\frac{m}{s}$ . It encounters a tailwind traveling at $9\frac{m}{s}$ . At what speed is the airplane traveling?

Possible Answers:

$37\frac{m}{s}$

$9\frac{m}{s}$

$28\frac{m}{s}$

$46\frac{m}{s}$

Correct answer:

$46\frac{m}{s}$

Explanation:

A tailwind is a wind traveling in the same direction as an object. The find the new speed of the airplane, we add the speed of the tailwind to the speed of the airplane.

$\vec{v}= (37+9 )\frac{m}{s}=46\frac{m}{s}$

Report an Error

Example Question #32 : Motion In One Dimension

A ball is dropped at time t=0 from 20 meters above the ground.

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

How long does it take for the ball to hit the ground?

Possible Answers:

-2s

Correct answer:

Explanation:

Using the following kinematic equation:

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

is the final position, in meters, ( for ground).

$x_{0}$ is the starting position ( 20m above ground)

$v_{0}$ is the starting velocity $(0\frac{m}{s})$ because the ball is dropped and therefor starts from rest).

is the acceleration due to gravity given by the problem statement.

Input these values into the first ruling equation and note that in this notation up is the positive direction so the acceleration due to gravity is negative:

$0m=20m+(0\frac{m}{s})(t)+\frac{1}{2}(-10\frac{m}{s^{2}})t^{2}$

Then we solve for .

$0=20+0-5t^{2}$

$-20=-5t^{2}$

$4=t^{2}$

t=2s

Report an Error

Example Question #31 : Motion In One Dimension

A hot air balloon is rising at $2 \frac{m}{s}$ when a bag falls from the edge of the basket.

How far has the bag fallen from where it left the basket after 3 seconds?

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

Possible Answers:

49m

30m

39m

45m

26m

Correct answer:

39m

Explanation:

We use the equation:

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

With $x_{0}=0m$ , $v_{0}=2\frac{m}{s}$ , $a=-10\frac{m}{s^{2}}$ , and t=3s we have:

$x=0+3*2+\frac{1}{2}*-10*3^{2}$

x=-39m

The bag has fallen 39 meters from its original position. Note that the question does not ask for the distance the bag is from the basket after 3 seconds so we do not have to take in to account how far the balloon has moved in the 3 seconds. Also, note that the question asks "how far" the bag is from the basket, so we do not have to worry about sign in the answer.

Report an Error

Example Question #31 : Motion In One Dimension

A rocket ship moves at $200\frac{m}{s}$ when a part breaks off and falls straight down. The rocket continues to move at the same velocity after the part breaks off.

How far below the ship is the part after 5 seconds?

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

Possible Answers:

1000m

125m

875m

75m

275m

Correct answer:

125m

Explanation:

We use the equation:

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

With $x_{0}=0m$ , $v_{0}=200\frac{m}{s}$ , $a=-10\frac{m}{s^{2}}$ , and t=5s we have:

$x=0+200*5+\frac{1}{2}*-10*5^{2}$

x=875m

The part moves 875 meters upward in the time it breaks from the ship. Then, we use:

x=vt with $v=200\frac{m}{s}$ and t=5s to find that the ship has moved 1000 meters during the same 5 second period.

So the ship moved upward 1000 meters in the 5 second time period while the part moved 875 meters upward so to find the distance between the ship and the part as the problem statement requested we subtract:

1000m-875m=125m

Report an Error

Example Question #33 : Motion In One Dimension

A car moving to the right at $20\frac{m}{s}$ accelerates to the right at $3\frac{m}{s^{2}}$ for 14 meters.

What is the cars velocity after the acceleration period?

Possible Answers:

$22\frac{m}{s}$

$29\frac{m}{s}$

$20\frac{m}{s}$

$18\frac{m}{s}$

$26\frac{m}{s}$

Correct answer:

$22\frac{m}{s}$

Explanation:

We use the equation:

$v_{f}^{2}=v_{0}^{2}+2ax$

Where the initial velocity is $v_{o}=20\frac{m}{s}$ , the acceleration is $a=3\frac{m}{s^{2}}$ , and the displacement during the acceleration x=14m .

Plug in those values into our equation and solve.

$v_f^2=(20\frac{m}{s})^2+2*(3\frac{m}{s^2})*14m$

$v_f=\sqrt{400+84}=\sqrt{484}=22\frac{m}{s}$

Report an Error

Example Question #34 : Motion In One Dimension

Calculate the acceleration, of a race car that can go a distance of 1320ft in 3.5s , assuming the acceleration is constant and the car starts from rest.

$1ft\approx 0.3048m$

Possible Answers:

$215.5 \frac{m}{s^2}$

$5.4\frac{m}{s^2}$

$229.7\frac{m}{s^2}$

$65.7 \frac{m}{s^2}$

Correct answer:

$65.7 \frac{m}{s^2}$

Explanation:

However unrealistic, if the acceleration of the race car were constant, we could calculate a value. We need everything in SI units, so we have to convert the distance to

$1320ft\left(\frac{0.3048m}{1ft}\right )=402.3m$

Using a kinematic equation, we can directly calculate the acceleration ,

$\Delta x = \frac{1}{2} a t^2 \Rightarrow a = \frac{2 \Delta x}{t^2}=\frac{2(402.3m)}{(3.5s)^2}=65.7\frac{m}{s^2}$

Report an Error

Example Question #2 : Calculating Motion In One Dimension

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:

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

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

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

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

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

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 #37 : Motion In One Dimension

Object A is traveling at $<10,0>\frac{m}{s}$ . It is hit from behind by object B traveling at $<15,0>\frac{m}{s}$ . If they both have a mass of 25kg , determine their final velocity if they stick together.

Possible Answers:

$<0,12.5>\frac{m}{s}$

$<15.0,0>\frac{m}{s}$

$<15,15>\frac{m}{s}$

$<0,0>\frac{m}{s}$

$<12.5,0>\frac{m}{s}$

Correct answer:

$<12.5,0>\frac{m}{s}$

Explanation:

$\overrightarrow{P}_{Total}=\overrightarrow{P}_1+\overrightarrow{P}_2...$

Definition of momentum:

$\overrightarrow{P}=m\overrightarrow{v}$

Combine equations:

$\overrightarrow{P}_{Total}=m_1\overrightarrow{v}_1+m_2\overrightarrow{v}_2$

Plug in values:

$\overrightarrow{P}_{Total}=25*<10,0>+25*<15,0>$

$\overrightarrow{P}_{Total}=<250,0>+<375,0>$

$\frac{\overrightarrow{P}_{Total}}{m}=\frac{<625,0>kg\frac{m}{s}}{50kg}$

$\overrightarrow{v}=<12.5,0>\frac{m}{s}$

Report an Error

Example Question #311 : Ap Physics 1

A ball is dropped from a building of height 45m. How long does it take for the ball to hit the ground? Ignore air resistance.

Possible Answers:

10s

Correct answer:

Explanation:

Let's first list the values we know.

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

$v_o=0\frac{m}{s}$

d=45m

We are looking for the time it takes for this ball to reach the ground, thus we are looking for . The kinematics equation we need to use is:

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

Since the initial velocity is zero, the v_ot term disappears. Thus we are left with:

$45m=\frac{1}{2}*10\frac{m}{s^2}t^2$

Solve for time.

t=3s

We only take the positive value in our calculation since time cannot be negative.

Report an Error

View AP Physics 1 Tutors

Muhammad
Certified Tutor

The University of Texas at Dallas, Bachelor of Science, Mechanical Engineering.

View AP Physics 1 Tutors

Shakeel
Certified Tutor

View AP Physics 1 Tutors

Siddharth
Certified Tutor

University of Petroleum and Energy Studies and, Masters in Business Administration, Aviation Management.

All AP Physics 1 Resources

7 Diagnostic Tests 170 Practice Tests Question of the Day Flashcards Learn by Concept

Popular Subjects

Reading Tutors in Miami, SAT Tutors in Miami, Spanish Tutors in Boston, Computer Science Tutors in Chicago, Computer Science Tutors in Dallas Fort Worth, MCAT Tutors in Philadelphia, ISEE Tutors in Washington DC, Algebra Tutors in Washington DC, Calculus Tutors in Houston, Computer Science Tutors in Philadelphia

Popular Courses & Classes

ACT Courses & Classes in San Francisco-Bay Area, GMAT Courses & Classes in Phoenix, MCAT Courses & Classes in Houston, Spanish Courses & Classes in Phoenix, ACT Courses & Classes in Chicago, LSAT Courses & Classes in New York City, GRE Courses & Classes in Denver, ACT Courses & Classes in Dallas Fort Worth, GRE Courses & Classes in New York City, ISEE Courses & Classes in New York City

Popular Test Prep

SSAT Test Prep in Phoenix, ACT Test Prep in Dallas Fort Worth, SSAT Test Prep in Washington DC, ISEE Test Prep in Philadelphia, MCAT Test Prep in Denver, SSAT Test Prep in Boston, GRE Test Prep in Boston, GRE Test Prep in Denver, GMAT Test Prep in Phoenix, ACT Test Prep in San Francisco-Bay Area

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 1 : Linear Motion and Momentum

Study concepts, example questions & explanations for AP Physics 1

All AP Physics 1 Resources

Example Questions

Example Question #31 : Linear Motion And Momentum

Example Question #31 : Linear Motion And Momentum

Example Question #32 : Motion In One Dimension

Example Question #31 : Motion In One Dimension

Example Question #31 : Motion In One Dimension

Example Question #33 : Motion In One Dimension

Example Question #34 : Motion In One Dimension

Example Question #2 : Calculating Motion In One Dimension

Example Question #37 : Motion In One Dimension

Example Question #311 : Ap Physics 1

All AP Physics 1 Resources

Report an issue with this question

DMCA Complaint

Contact Information

Address

Complaint Details

Email address:
Your name:
Feedback: