Call Now to Set Up Tutoring:

(888) 888-0446

High School Physics : Linear Motion

Study concepts, example questions & explanations for High School Physics

Create An Account Create Tests & Flashcards

所有高中物理资源

6 Diagnostic Tests 233 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

← Previous 1 2 … 8 9 10 11 12 13 14 15 16 Next →

Example Question #141 :Linear Motion

A car starts from rest, speeds up with constant acceleration and travels 100m in 5s. What is the final velocity and the acceleration of the car?

Possible Answers:

$V_{f}=24\frac{m}{s};a=8\frac{m}{s^2}$

$V_{f}=25\frac{m}{s};a=15\frac{m}{s^2}$

$V_{f}=32\frac{m}{s};a=6.7\frac{m}{s^2}$

$V_{f}=40\frac{m}{s};a=8\frac{m}{s^2}$

$V_{f}=50\frac{m}{s};a=10\frac{m}{s^2}$

Correct answer:

$V_{f}=40\frac{m}{s};a=8\frac{m}{s^2}$

Explanation:

The easiest way to approach this problem is find the average velocity, multiply by 2 because the car started from rest, and then divide the final velocity by time to get the acceleration.

Average velocity:

$V_{avg}=\frac{d}{t}=\frac{100m}{5s}=20\frac{m}{s}$

Final velocity:

The final velocity is $40\frac{m}{s}$ because the car started from rest. This is evident from the equation for average velocity if we solve for final velocity:

$V_{avg}=\frac{V_{0}+V_{f}}{2}$

$V_0=2V_{avg}-V_f$

$V_0=2*40-40=40\frac{m}{s}$

Acceleration:

$a=\frac{V_0-V_f}{t}=\frac{40\frac{m}{s}}{5s}=8\frac{m}{s^2}$

Report an Error

Example Question #142 :Linear Motion

Trevor is traveling 20m/s when he sees a red light ahead. His car is capable of decelerating at a rate of 3.45m/s^2 . If it takes him 0.360s to get the brakes on and he is 19.0m from the intersection when he sees the light, how far from the beginning of the intersection will he be, and in what direction? In other words, will he be able to stop in time?

Possible Answers:

No, he will not be able to stop in time and will stop about 15.8 m beyond the intersection.

Yes, he will be able to stop in time and will stop about 13.3 m before the intersection.

No, he will not be able to stop in time and will stop about 46.2 m beyond the intersection.

Yes, he will be able to stop in time and will stop about 5.7m before the intersection.

Correct answer:

No, he will not be able to stop in time and will stop about 46.2 m beyond the intersection.

Explanation:

Knowns:

v_0 = 20m/s

v_f=0m/s

a = -3.45m/s^2

t_0 = 0s

$t_{reaction} = 0.360s$

$\Delta x = 19m$

Unknowns:

$\Delta x$ that Trevor actually travels = ?

Equation:

Since Trevor takes a moment to react before stepping on the breaks, determine how far Trevor travels during the reaction time. At this time he is traveling at a constant velocity.

$v = \frac{\Delta x}{\Delta t}$

$\Delta x = v*\Delta t$

$\Delta x = 20m/s * 0.360s$

$\Delta x = 7.2m$

Next, using kinematic equations to determine where Trevor stops his car based on the acceleration of his car.

$v^f_2=v^0_2 +2a\Delta x$

$\Delta x =\frac{v^f_2-v^0_2}{2a}$

$\Delta x =\frac{0^2-(20m/s)^2}{2(-3.45m/s^2)}$

$\Delta x = 57.97m$ beyond the reaction point

Total distance traveled is the reaction time distance plus the distance traveled during the deceleration period.

total distance traveled = 7.2m +57.97m = 65.17m

To find the distance beyond the red light, subtract the distance traveled from the distance to the light

distance beyond red light = 65.17m-19m = 46.2m

Report an Error

Example Question #143 :Linear Motion

Along a highway there is an unmarked police car traveling a constant 90km/h . The officer is passed by a speeder traveling 130km/h . Preciselyafter the speeder passes, the officer steps on the accelerator to catch the speeder. If the police car’s acceleration is 2.50m/s^2 , how much time passes before the police car overtakes the speeder? Assume here that the speeder is moving at a constant speed.

Possible Answers:

16.00 s

11.11 s

Not enough information is provided

4.40 s

9.79 s

Correct answer:

9.79 s

Explanation:

Knowns:

$v_{speeder} = 130km/h$

$v_{officer} =900km/h$

$t_{0 speeder} = 0 seconds$

$t_{0 officer}= 1 second$

a=2.50m/s^2

Unknowns:

$t_{final}=?$

Equation:

To make things easier, set the frame of reference to the police officer. This means that the police officer is traveling at 0km/h and the speeder is moving at a speed relative to the officer.

$v_{speeder in reference to officer} = v_{speeder}- v_{officer}$

v = 130km/h-90km/h

v= 40km/h

So the new relative velocities are:

$v_{speeder} = 40km/h$

$v_{officer} =0km/h$

This will help make the math easier.

Convert the relative velocity of the speeder to m/s.

$\frac{40km}{h}*\frac{1000m}{1km}*\frac{1h}{60 min}*\frac{1min}{60s}= 11.11m/s$

Now there are two equations that have to be true for the officer to catch up to the speeder.

$\Delta x_{officer} =v_o\Delta t_{officer} + \frac{1}{2}a\Delta t^2_{officer}$

$v_{speeder}=\frac{\Delta x_{speeder}}{\Delta t_{speeder}}$

There are two things that connect these equations. First, the distance that the officer travels and the speeder travels must be the same. Additionally the final velocity of both the officer and the speeder must be the same.

Rearrange the speeder equation for the distance traveled.

$v_{speeder} * \Delta t_{speeder} = \Delta x_{speeder}$

Set the two equations equal, as the distance traveled of the speeder is equal to the distance traveled by the officer.

$v_o\Delta t_{officer} + \frac{1}{2}a\Delta t^2_{officer}=v_{speeder} * \Delta t_{speeder}$

记住,参考点是在办公室r’s perspective so the officer has an initial velocity of 0m/s.

$0m/s + \frac{1}{2}a\Delta t^2_{officer}=v_{speeder} * \Delta t_{speeder}$

Rewrite the equation breaking up the change of time into time final and time initial.

$\frac{1}{2}a(t_{f officer}- t_0 officer)^2=v_{speeder} * (t_{f speeder} -t_{0 speeder})$

Distribute on the left side.

$\frac{1}{2}a(t^2_{f officer}- t_{0 officer}t_{f officer} +t^2_{0 officer})=v_{speeder} * (t_{f speeder} -t_{0 speeder})$

$\frac{1}{2}at^2_{f officer}- \frac{1}{2}at_{0 officer}t_{f officer} +\frac{1}{2}at^2_{0 officer}=v_{speeder} * (t_{f speeder} -t_{0 speeder})$

Substitute values

$\frac{1}{2}(2.5m/s^2)t^2_{f officer}- \frac{1}{2}(2.5m/s^2)(1s)t_{f officer} +\frac{1}{2}(2.5m/s^2) (1s)^2=11.11m/s* (t_{f speeder} -0s)$

Simplify

$(1.25)t^2_{f officer}- 1.25t_{f officer} +1.25=11.11 t_{f speeder}$

Remember that the final time for both the officer and the speeder are the same. So rearrange this in the form of a quadratic equation.

$(1.25)t^2_{f officer}- 12.36t_{f officer} +1.25=0$

Use the quadratic formula to solve.

Possible answers: t= 0.10s or t = 9.79s

Since the officer did not move until, then the second answer must be correct.

Report an Error

Example Question #144 :Linear Motion

Usain Bolt accelerates from rest to 13.4m/s over a distance of 100m . What is his acceleration? Assume the acceleration is constant.

Possible Answers:

a=.879m/s^2

a=9.8m/s^2

a=.847m/s^2

a=.898m/s^2

a=.869m/s^2

Correct answer:

a=.898m/s^2

Explanation:

We are given final velocity and we know initial velocity is 0. We can use the following kinematics formula to relate final speed, initial speed, acceleration, and displacement:

$v_f^2=v_0^2+2a\Delta x$

The initial velocity is zero since Usain starts from rest. Therefore we can remove it from the equation.

$v_f^2=2a\Delta x$

Rearrange the equation to solve for acceleration.

$\frac{v_f^2}{2\Delta x}=a$

$\frac{(13.4m/s)^2}{2(100m)}=a$

a = .898m/s^2

Report an Error

Example Question #145 :Linear Motion

Peter starts from rest and runs down a hallway 31 seconds . If his final velocity is 13m/s , how far did he run?

Possible Answers:

222m

13.3m

0.387m

100.75m

201.8m

Correct answer:

201.8m

Explanation:

Knowns:

v_0 = 0m/s

v_f =13m/s

$\Delta t =31s$

Unknowns:

a = ?

$\Delta x = ?$

Equation:

First we need to determine Peter’s acceleration. The best kinematic equation to use here is:

$v_f=v_0+a\Delta t$

$\frac{v_f-v_0}{\Delta t}=a$

$\frac{13m/s-0m/s}{31s}=a$

a = 0.42m/s^2

Once Peter’s acceleration is known it is possible to find the distance he traveled.

$\Delta x = v_0 \Delta t+\frac{1}{2}a\Delta t^2$

Peter’s initial velocity is 0m/s so this equation simplifies to

$\Delta x =\frac{1}{2}a\Delta t^2$

$\Delta x =\frac{1}{2}(0.42m/s^2)(31)^2$

$\Delta x = 201.8m$

Report an Error

Example Question #146 :Linear Motion

Objectsandboth start from rest. They both accelerate at the same rate. However, objectaccelerates for twice the time as object. What is the distance traveled by objectcompare to that of object?

Possible Answers:

The same distance

Four times as far

Three times as far

Twice as far

Correct answer:

Four times as far

Explanation:

$\Delta x = v_0 \Delta t+\frac{1}{2}a\Delta t^2$

The distance traveled is directly related to the square of the time traveled. Therefore time if time is doubled, the value will be squared and therefore four times as great. The distance traveled will be 4 times as far.

Report an Error

Example Question #147 :Linear Motion

Suppose a can is kicked and then travels up a smooth hill of ice. Which of the following is true about its acceleration?

Possible Answers:

It will have a constant acceleration up the hill, but a different constant acceleration when it comes back down

It will have a varying acceleration along the hill.

It will travel at constant velocity with zero acceleration

It will have the same acceleration, both up the hill and down the hill.

Correct answer:

It will have the same acceleration, both up the hill and down the hill.

Explanation:

The can will have the same acceleration both up and down the hill since there is no friction. Friction or other forces would cause a change in acceleration. But since there is no friction, the can will travel up the hill, slow down and then accelerate back down the hill.

Report an Error

Example Question #148 :Linear Motion

A runner wants to complete a 10,000m run in less than 30 minutes . After running at constant speed for exactly 26 minutes , the runner still has 1500 meters left to run. The runner must then accelerate at 0.30m/s^2 for how many seconds in order to reach a final velocity that will allow them to complete theleft of the race in the desired time?

Possible Answers:

8.32 s

20.83 s

1.48 s

6.70 s

2.67 s

Correct answer:

2.67 s

Explanation:

Knowns:

$\Delta x_{total} = 10,000m$

$\Delta t_{total} = 30 min$

$\Delta t_{constant speed} = 26 min$

$\Delta x_{accelerating} = 1500m$

a = 0.30m/s^2

Unknown:

$\Delta t_{accelerating}= ?$

Equation:

The first thing is to determine the initial velocity of the runner before the runner accelerates for the final portion of the race. Since the runner is traveling at a constant velocity

$v = \frac{\Delta x}{\Delta t}$

Next, convert the time during the constant speed portion to seconds.

$26 min *\frac{60s}{1 min}= 1560s$

Determine the amount of distance traveled while at a constant speed.

10,000m-1,500m = 8,500m

Use these values to determine the velocity of the runner.

$v = \frac{8,500m}{1560s}$

v = 5.45m/s

Next determine the final velocity needed for the runner to finish out the race in the remaining time.

30min - 26min = 4 min left in race

$4min *\frac{60s}{1 min} = 240s$

$v = \frac{1500m}{240s}$

v = 6.25 m/s

Finally use the kinematic equations to calculate the time needed to get to this velocity.

$v_f= v_0+a\Delta t$

Rearrange for time

$\Delta t = \frac{v_f-v_0}{a}$

$\Delta t = \frac{6.25m/s-5.45m/s}{0.30m/s^2}$

$\Delta t = 2.67s$

Report an Error

Example Question #149 :Linear Motion

A ball rolls to a stop after 4.2s . If it had a starting velocity of 8.1ms , what is the deceleration on the ball due to friction?

Possible Answers:

6.93m/s^2

3.9m/s^2

-4.2m/s^2

-0.526m/s^2

-1.93m/s^2

Correct answer:

-1.93m/s^2

Explanation:

We are given the initial velocity, time, and final velocity (zero because the ball stops). Using these values and the appropriate motion equation, we can solve for the acceleration.

Acceleration is given by the change in velocity over time:

$a = \frac{v_f-v_0}{\Delta t}$

We can use our values to solve for the acceleration.

$a = \frac{0m/s - 8.1m/s}{4.2s}$

a = -1.93m/s^2

Report an Error

Example Question #150 :Linear Motion

Objectsand同时开始休息。他们都在sa加速me rate. However, objectaccelerates for twice the time as object. What is the final speed of objectcompared to that of object?

Possible Answers:

Four times as fast

The same speed

Twice as fast

Three times as fast

Correct answer:

Twice as fast

Explanation:

$v_f = v_0+a\Delta t$

There is a direct relationship between the final velocity and the time traveled. Therefore if the time is doubled, the velocity would double as well.

Report an Error

← Previous 1 2 … 8 9 10 11 12 13 14 15 16 Next →

View Tutors

Terry
Certified Tutor

Yale University, Bachelor in Arts, Latin American Studies. Princeton Theological Seminary, Master of Divinity, Theology.

View Tutors

Phyllis
Certified Tutor

University of Michigan-Ann Arbor, Bachelor in Arts, Arts, General.

View Tutors

Ketrina
Certified Tutor

Bethune-Cookman University, Bachelor in Arts, Education. Bethune-Cookman University, Bachelor of Fine Arts, Theater Arts.

所有高中物理资源

6 Diagnostic Tests 233 Practice Tests Question of the Day Flashcards Learn by Concept

Popular Subjects

Math Tutors in Dallas Fort Worth,Reading Tutors in San Diego,Calculus Tutors in Denver,Statistics Tutors in San Diego,Spanish Tutors in San Francisco-Bay Area,SSAT Tutors in San Diego,代数Tutors in Denver,坐在导师在凤凰城,GMAT Tutors in Philadelphia,LSAT Tutors in Washington DC

Popular Courses & Classes

Spanish Courses & Classes in Phoenix,LSAT Courses & Classes in Seattle,GRE Courses & Classes in Chicago,ACT Courses & Classes in New York City,ACT Courses & Classes in Miami,MCAT Courses & Classes in Philadelphia,MCAT Courses & Classes in Miami,MCAT Courses & Classes in Seattle,MCAT Courses & Classes in Boston,ISEE Courses & Classes in Seattle

Popular Test Prep

MCAT Test Prep in Houston,SSAT Test Prep in Philadelphia,GRE Test Prep in San Francisco-Bay Area,LSAT Test Prep in Atlanta,ACT Test Prep in New York City,SSAT Test Prep in Dallas Fort Worth,MCAT Test Prep in Denver,GMAT Test Prep in Los Angeles,MCAT Test Prep in Phoenix,MCAT 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

*电话号码

*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

High School Physics : Linear Motion

Study concepts, example questions & explanations for High School Physics

所有高中物理资源

Example Questions

Example Question #141 :Linear Motion

Example Question #142 :Linear Motion

Example Question #143 :Linear Motion

Example Question #144 :Linear Motion

Example Question #145 :Linear Motion

Example Question #146 :Linear Motion

Example Question #147 :Linear Motion

Example Question #148 :Linear Motion

Example Question #149 :Linear Motion

Example Question #150 :Linear Motion

所有高中物理资源

Report an issue with this question

DMCA Complaint

Contact Information

Address

Complaint Details

Find the Best Tutors

Email address:
Your name:
Feedback: