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 #571 : Ap Physics 1

$27.8\frac{m}{s}$ is the minimum speed needed for an airplane to take off. The airplane is accelerating at a constant $3.3\frac{m}{s^{2}}$ . What length does the runway need to be?

Possible Answers:

386m

111 m

117 m

772m

400 m

Correct answer:

117 m

Explanation:

The correct answer is 117 m . Because we don't have time, we should use an equation that does not require a time variable, like

$v^{2}=v_{o}^{2} + 2ax$

where is the final velocity, $v_{o}$ is the initial velocity, is the acceleration, and is the distance traveled. We are given , $v_{o}$ , and . Therefore, we can put these values into our equation to solve for :

$(27.8\frac{m}{s})^{2} = (0\frac{m}{s})^{2} + (2)(3.3\frac{m}{s^{2}})x$

$772.8 \frac{m}{s}= (6.6\frac{m}{s^{2}})x$

x=117m

Report an Error

Example Question #573 : Ap Physics 1

A car is moving with a velocity given by $v(t)=3t^2+5t+4\ m/s$ after being ejected by a connon at an initial velocity of $100\ m/s$ .

When is the acceleration equal to $125\frac{m}{s^2}$ ?

Possible Answers:

24s

27.5s

17.5 s

20 s

15 s

Correct answer:

20 s

Explanation:

In order to find the acceleration, you must first take the derivative of the velocity:

$v(t)=3t^2+5t+4\frac{m}{s}$

$a(t)=v'(t)=6t+5\frac{m}{s^2}$

The initial velocity does not affect the acceleration! It would only affect the position function! Afterwards, just plug in and solve for t:

6t+5=125

6t=120

t=20s

Report an Error

Example Question #573 : Ap Physics 1

You drive 3 mi east from your home past the grocery store. As you pass the store, you drive another 4 mi eastward to the post office. From the post office, you drive 8 mi westward. Then, you drive 1 mi east back home. Assume you do not stop and are traveling at a constant speed. The trop takes you 30 min total. What was your average speed? Your average velocity? (average speed, average velocity)

Possible Answers:

$32\frac{m}{s}$ , $0\frac{m}{s}$

$14.3 \frac{m}{s}$ , $0\frac{m}{s}$

$0.533\frac{m}{s}$ , $0\frac{m}{s}$

$32\frac{m}{s}$ , $32\frac{m}{s}$

$14.3\frac{m}{s}$ , $14.3\frac{m}{s}$

Correct answer:

$14.3 \frac{m}{s}$ , $0\frac{m}{s}$

Explanation:

The correct answer is $14.3\frac{m}{s}$ , $0\frac{m}{s}$ . Average speed is:

$\frac{distance}{time}$ , in $\frac{meters(m)}{second(s)}$

We traveled a total of 16 mi . Convert this to meters to get 25749.5 m

Then we must convert 30 min to

$30 min (\tfrac{60 s}{1 min})=1800 s$

Then divide to get your average speed. This gives us $14.3\frac{m}{s}$

Velocity, on the other hand, is $\frac{displacement}{time}$ in $\frac{m}{s}$ .

Displacement is the the final distance from the initial position. Displacement is considered a state function. Only the final position matters; it doesn't matter how you get there. For example, if you start at point A and drive in a circle that covers 100 mi , and end up at point A again, your displacement is 0 m . This is because your final position is the same as your initial position. So in this problem, because your final position is the same as your initial position, your displacement is 0 m .

So $\frac{0 m}{1800 s}=0\frac{m}{s}$ for the average velocity.

Report an Error

Example Question #574 : Ap Physics 1

On your way home from school, you drive 3 mi until you pass John's house. Once you pass John's house you drive another 2 mi in the same direction until you reach the park, where you have soccer practice. The trip takes you 7 min . What is your average speed? Your average velocity? (Average speed, average velocity).

Possible Answers:

$22.4\frac{m}{s}$ , $22.4\frac{m}{s}$

$13.4\frac{m}{s}$ , $13.4\frac{m}{s}$

$0.833\frac{m}{s}$ , $0.833\frac{m}{s}$

$14.0\frac{m}{s}$ , $12.0\frac{m}{s}$

$22.4\frac{m}{s}$ , $0\frac{m}{s}$

Correct answer:

$22.4\frac{m}{s}$ , $22.4\frac{m}{s}$

Explanation:

The correct answer is $22.4\frac{m}{s}$ , $22.4\frac{m}{s}$

Remember average speed is $\frac{distance}{time}$ , also $\frac{meters(m)}{second(s)}$

By converting miles to meters and minutes to seconds , we get $\frac{8046m}{360s}=22.4 \frac{m}{s}$

This is also true for average velocity. Average velocity is $\frac{displacement}{time}$ , $\frac{meters(m)}{second(s)}$

Displacement is the distance of the final position from the initial position. Because we do not change direction, the displacement is equal to the distance. That means that the average velocity will be equal to the average speed (time remains constant).

Report an Error

Example Question #31 : Fundamentals Of Displacement, Velocity, And Acceleration

A car drives 100 m in 1 m at constant velocity and without changing direction. What is the car's acceleration?

Possible Answers:

$0\frac{m}{s^{2}}$

$100\frac{m}{s^{2}}$

$7.0\frac{m}{s^{2}}$

$16.7\frac{m}{s^{2}}$

$1.67\frac{m}{s^{2}}$

Correct answer:

$0\frac{m}{s^{2}}$

Explanation:

Average acceleration is equal to $\frac{(velocity_{2}-velocity_{1})}{time}$ , where $V_{2}$ is the final velocity and $V_{1}$ is the initial velocity.

If the velocity is constant, the final and initial velocities are equal. Let's say the car is traveling at $30\frac{m}{s}$ constantly.

$30\frac{m}{s}- 30\frac{m}{s}=0\frac{m}{s}$ . So $\frac{0\frac{m}{s}}{\frac{60}{s}}= 0\frac{m}{s^{2}}$

When velocity is constant, acceleration is always equal to $0\frac{m}{s^{2}}$

Report an Error

Example Question #32 : Fundamentals Of Displacement, Velocity, And Acceleration

A car accelerates to $75\frac{km}{h}$ from rest in 5 s . Assume the car moves in a straight line. What is the magnitude of the car's acceleration?

Possible Answers:

$4.17\frac{m}{s^{2}}$

$14.3\frac{m}{s^{2}}$

$3.32\frac{m}{s^{2}}$

$20.8\frac{m}{s^{2}}$

$15.0\frac{m}{s^{2}}$

Correct answer:

$4.17\frac{m}{s^{2}}$

Explanation:

Average acceleration is equal to $\frac{(velocity_{2}-velocity_{1})}{time}$ where $V_{2}$ is the final velocity and $V_{1}$ is the initial velocity.

In this problem, we know our initial velocity is $0\frac{km}{h}$ (starting from rest). The final velocity is $75\frac{km}{h}$ . We must convert these units to $\frac{meters}{second}$ .

$75 km\frac{1000m}{1 km}=75000m$

$60 min\frac{60 s}{1 min}=3600s$

Using these values, we can calculate the final velocity to be $\frac{75000m}{3600s}=20.8\frac{m}{s}$

Using this number, we can finally calculate the magnitude of acceleration.

$({20.8\frac{m}{s}}- 0\frac{m}{s})(5s)=4.17\frac{m}{s^{2}}$

Report an Error

Example Question #33 : Fundamentals Of Displacement, Velocity, And Acceleration

A car at a stop sign must cross an intersection that is 20.0m long. If the car accelerates at $2.5\frac{m}{s^2}$ constantly, how long will it take for the car to cross the intersection?

Possible Answers:

8.0s

4.47s

4.0s

16.0s

9.0s

Correct answer:

4.0s

Explanation:

We can solve this problem using the kinematic equation:

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

Where is the displacement, v_o is the initial velocity, is the acceleration, and is time. We are given values for all but the time. Therefore, we may plug these values into the above equation and solve for .

$20.0m=\left(0\frac{m}{s}*t\right)+\left(\frac{1}{2}*2.5\frac{m}{s}^2*t^2\right)$

20.0 m=t^2

t=4.0s

Report an Error

Example Question #34 : Fundamentals Of Displacement, Velocity, And Acceleration

A ball is thrown upward at $3.0\frac {m}{s}$ from a tower. Ignore air resistance. How far does the ball travel after 1 s ?

Possible Answers:

7.9 m

6.8 m

-1.9 m

-6.8 m

1.9m

Correct answer:

-1.9 m

Explanation:

The correct answer is -1.9m . We can use the kinematics equation:

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

We are given the values of $v_{o}$ , , and (initial velocity, time, and acceleration). Therefore we can solve for to find the distance traveled by the ball.

$y= (3\frac{m}{s})(1s) + (\frac{1}{2})(9.8\frac{m}{s^{2}})(1.0s)^{2}$

y= (3m)+(-4.9m)

y=-1.9m

is negative because the position of the ball is 1.9m below its initial point.

Report an Error

Example Question #35 : Fundamentals Of Displacement, Velocity, And Acceleration

From the top of a mountain, you throw a rock downward at $3.0\frac{m}{s}$ . What is the rock's displacement after 2 s ?

Possible Answers:

19.6 m

-25.6 m

-15.8m

-19.6 m

15.8 m

Correct answer:

-25.6 m

Explanation:

To solve this problem, we can use the kinematics equation:

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

Where $v_{o}$ is the initial velocity, is time, and is acceleration. We are given all of these variables, so we can plug them in for .

$y = (-3.0\frac{m}{s})(2.0s) + (\frac{1}{2})(-9.8\frac{m}{s^{2}})(2.0s)^{2}$

y=-25.6m

In this problem, $v_{o}$ and are negative because the rock is initially being thrown downhill and is accelerating downwards.

Report an Error

Example Question #36 : Fundamentals Of Displacement, Velocity, And Acceleration

If the driver of a $1200\ kg$ car traveling at $32\ \frac{m}{s}$ slams on the brakes, and it takes $2.0\ s$ for the car to come to a complete stop, what is the coefficient of kinetic friction between the car's tires and the asphalt that the car is traveling on?

Possible Answers:

0.54

0.85

2.4

1.2

1.6

Correct answer:

1.6

Explanation:

The first step in doing this problem is figuring out how many forces are acting on the car when the driver slams the brakes. Since there is nothing pushing the car in the direction that it was originally traveling, the only force that is acting on the car is the force of friction. This means that the net force is simply just the force of kinetic friction, net force is also equal to mass multiplied by acceleration, so this equation appears as follows:

$F_{k}=ma$

where $F_{k}$ is the force of kinetic friction.

The force of friction is also equal to normal force multiplied by the coefficient of kinetic friction for a certain surface. This equation appears as follows:

$F_{k}=\mu _{k}N$

where $F_{k}$ is the force of kinetic friction, $\mu _{k}$ is the coefficient of kinetic friction, and N is normal force (also equal to mass x gravity).

These 2 equations can be connected in order to figure out the coefficient of kinetic friction.

$ma=\mu _{k}mg$

mass can be cancelled from both sides. Acceleration is found as follows:

$a=\frac{\Delta V}{\Delta t}=\frac{32\ \frac{m}{s}}{2\ s}=16\ \frac{m}{s^{2}}$

By plugging in the given values, $\mu _{k}$ is found to be:

$16=\mu _{k}*9.81$

$\mu _{k}=1.63$

Report an Error

View AP Physics 1 Tutors

Jethro
Certified Tutor

Federal University of Agriculture Abeokuta,, Bachelor of Engineering, Mechatronics, Robotics, and Automation Engineering.

View AP Physics 1 Tutors

Muhammad
Certified Tutor

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

View AP Physics 1 Tutors

Aaron
Certified Tutor

Tulsa Community College, Associate in Science, Mechanical Engineering.

All AP Physics 1 Resources

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

Popular Subjects

MCAT Tutors in Denver, French Tutors in Atlanta, Algebra Tutors in San Diego, GRE Tutors in Houston, French Tutors in Boston, Biology Tutors in New York City, Chemistry Tutors in Washington DC, English Tutors in San Diego, Biology Tutors in Dallas Fort Worth, MCAT Tutors in Philadelphia

Popular Courses & Classes

SAT Courses & Classes in Phoenix, SAT Courses & Classes in Miami, ISEE Courses & Classes in Phoenix, SAT Courses & Classes in Denver, ISEE Courses & Classes in Seattle, GMAT Courses & Classes in San Diego, ACT Courses & Classes in Atlanta, SSAT Courses & Classes in Denver, GMAT Courses & Classes in Phoenix, SSAT Courses & Classes in Chicago

Popular Test Prep

GRE Test Prep in San Diego, LSAT Test Prep in Phoenix, ISEE Test Prep in Seattle, ACT Test Prep in Seattle, MCAT Test Prep in Washington DC, MCAT Test Prep in Denver, GRE Test Prep in Houston, GMAT Test Prep in San Diego, SSAT Test Prep in Seattle, LSAT Test Prep in Philadelphia

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 #571 : Ap Physics 1

Example Question #573 : Ap Physics 1

Example Question #573 : Ap Physics 1

Example Question #574 : Ap Physics 1

Example Question #31 : Fundamentals Of Displacement, Velocity, And Acceleration

Example Question #32 : Fundamentals Of Displacement, Velocity, And Acceleration

Example Question #33 : Fundamentals Of Displacement, Velocity, And Acceleration

Example Question #34 : Fundamentals Of Displacement, Velocity, And Acceleration

Example Question #35 : Fundamentals Of Displacement, Velocity, And Acceleration

Example Question #36 : Fundamentals Of Displacement, Velocity, And Acceleration

All AP Physics 1 Resources

Report an issue with this question

DMCA Complaint

Contact Information

Address

Complaint Details

Email address:
Your name:
Feedback: