Call Now to Set Up Tutoring:

(888) 888-0446

Calculus AB : Calculate Position, Velocity, and Acceleration

Study concepts, example questions & explanations for Calculus AB

Create An Account

All Calculus AB Resources

45 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

Example Question #1 : Calculate Position, Velocity, And Acceleration

The position of a car is given by the following function:

$f(x)=\cos(x)e^{11x}+\csc(x)$

What is the velocity function of the car?

Possible Answers:

$\sin(x)e^{11x}+11\cos(x)e^{11x}-\csc(x)\cot(x)$

$-\sin(x)e^{11x}+11\cos(x)e^{11x}+\csc(x)\cot(x)$

$-\sin(x)e^{11x}+11\cos(x)e^{11x}$

$-\sin(x)e^{11x}+11\cos(x)e^{11x}-\csc(x)\cot(x)$

$-\sin(x)e^{11x}+\cos(x)e^{11x}-\csc(x)\cot(x)$

Correct answer:

$-\sin(x)e^{11x}+11\cos(x)e^{11x}-\csc(x)\cot(x)$

Explanation:

The velocity function of the car is equal to the first derivative of the position function of the car, and is equal to

$-\sin(x)e^{11x}+11\cos(x)e^{11x}-\csc(x)\cot(x)$

The derivative was found using the following rules:

$\frac{\mathrm{d} }{\mathrm{d} x} f(x)g(x)=f'(x)g(x)+f(x)g'(x)$ , $\frac{\mathrm{d} }{\mathrm{d} x} e^u=e^u \frac{\mathrm{du} }{\mathrm{d} x}$ , $\frac{\mathrm{d} }{\mathrm{d} x} \cos(x)=-\sin(x)$ , $\frac{\mathrm{d} }{\mathrm{d} x} \csc(x)=-\csc(x)\cot(x)$

Report an Error

Example Question #1 : Contextual Applications Of Derivatives

Let

$f(x)=e^{\pi x}$

Find the first and second derivative of the function.

Possible Answers:

$f'(x)=e^{\pi }$

$f''(x)= e^{x}$

$f'(x)=\pi e^{\pi x }$

$f''(x)=\pi^2 e^{\pi x}$

$f'(x)=e^{\pi x }$

$f''(x)=e^{\pi x}$

$f'(x)=\pi$

$f''(x)=\0$

Correct answer:

$f'(x)=\pi e^{\pi x }$

$f''(x)=\pi^2 e^{\pi x}$

Explanation:

In order to solve for the first and second derivatives, we must use the chain rule.

The chain rule states that if

y=f(u)

and

u=g(x)

then the derivative is

$\frac{dy}{dx}=\frac{dy}{du}*\frac{du}{dx}$

In order to find the first derivative of the function

$f(x)=e^{\pi x}$

we set

y=e^u

and

$u=\pi x$

Because the derivative of the exponential function is the exponential function itself, we get

$\frac{dy}{du}=\frac{d}{du}[e^u]$

$\frac{dy}{du}=e^u$

And differentiating we use the power rule which states

$\frac{\mathrm{d} }{\mathrm{d} x}x^n=nx^{n-1}$

As such

$\frac{du}{dx}=\frac{\mathrm{d} }{\mathrm{d} x}[\pi x]$

$\frac{du}{dx}=\pi x^{1-1}=\pi$

And so

$\frac{dy}{dx}=e^{\pi x}*\pi=\pi e^{\pi x}$

$f'(x)=\pi e^{\pi x}$

To solve for the second derivative we set

$y=\pi e^{u}$

and

$u=\pi x$

Because the derivative of the exponential function is the exponential function itself, we get

$\frac{dy}{du}=\frac{d}{du}[\pi e^u]$

$\frac{dy}{du}=\pi e^u$

And differentiating we use the power rule which states

$\frac{\mathrm{d} }{\mathrm{d} x}x^n=nx^{n-1}$

As such

$\frac{du}{dx}=\frac{\mathrm{d} }{\mathrm{d} x}[\pi x]$

$\frac{du}{dx}=\pi x^{1-1}=\pi$

And so the second derivative becomes

$\frac{dy}{dx}=\pi e^{\pi x}*\pi=\pi^2 e^{\pi x}$

$f''(x)=\pi^2e^{\pi x}$

Report an Error

Example Question #1 : Calculate Position, Velocity, And Acceleration

Find the velocity function of the particle if its position is given by the following function:

$s(t)=e^t\tan(t)+t^5+3$

Possible Answers:

$e^t\tan(t)-e^t\sec^2(t)+5t^4$

$e^t\tan(t)+e^t\sec^2(t)+t^5$

$e^t\tan(t)+e^t\sec^2(t)+5t^4$

$te^t\tan(t)+e^t\sec^2(t)+5t^4$

Correct answer:

$e^t\tan(t)+e^t\sec^2(t)+5t^4$

Explanation:

The velocity function is given by the first derivative of the position function:

$v(t)=s'(t)=e^t\tan(t)+e^t\sec^2(t)+5t^4$

and was found using the following rules:

$\frac{\mathrm{d} }{\mathrm{d} x} f(x)g(x)=f'(x)g(x)+f(x)g'(x)$ , $\frac{\mathrm{d} }{\mathrm{d} x} x^n=nx^{n-1}$ , $\frac{\mathrm{d} }{\mathrm{d} x} \tan(t)=\sec^2(t)$ , $\frac{\mathrm{d} }{\mathrm{d} x} e^x=e^x$

Report an Error

Example Question #1 : Calculate Position, Velocity, And Acceleration

Find the first and second derivatives of the function

f(x)=sec(x)

Possible Answers:

f'(x)=cot(x)

f''(x)=csc^2(x)

f'(x)=sec^2(x)

f''(x)=sec^3(x)

f'(x)=sec(x)tan(x)

f''(x)=sec^3(x)+sec(x)tan^2(x)

f'(x)=csc(x)

f''(x)=-sec(x)

Correct answer:

f'(x)=sec(x)tan(x)

f''(x)=sec^3(x)+sec(x)tan^2(x)

Explanation:

We must find the first and second derivatives.

We use the properties that

The derivative of is
The derivative of is

As such

$f'(x)=\frac{\mathrm{d} }{\mathrm{d} x}[sec(x)]=sec(x)tan(x)$

To find the second derivative we differentiate again and use the product rule which states

$\frac{\mathrm{d} }{\mathrm{d} x}[uv]=uv'+vu'$

Setting

u=sec(x) and v=tan(x)

we find that

$f''(x)=\frac{\mathrm{d} }{\mathrm{d} x}[sec(x)tan(x)]=sec(x)*sec^2(x)+tan(x)*sec(x)tan(x)$

=sec^3(x)+sec(x)tan^2(x)

As such

f'(x)=sec(x)tan(x)

f''(x)=sec^3(x)+sec(x)tan^2(x)

Report an Error

Example Question #2 : Calculate Position, Velocity, And Acceleration

Given the velocity function

$v(x)=\int_{0}^{x^2+2x+1} z^d\sin(z^5) dz$

where is real number such that $d\in [0,1]$ , find the acceleration function

a(x) .

Possible Answers:

$a(x)=(2x+2)(x+1)^{2d}\sin (x+1)^{10}$

$a(x)=(2x+2)^{d}(x+1)^{2d}\sin (x+1)^{10}$

$a(x)=(x+1)^{d}\sin (x+1)^{10}$

$a(x)=(x^2+2x+1)^{d}\sin (x^2+2x+1)^{10}$

$a(x)=(x^2+2x+1)^{2d}\sin (x^2+2x+1)^{10}$

Correct answer:

$a(x)=(2x+2)(x+1)^{2d}\sin (x+1)^{10}$

Explanation:

We can find the acceleration function a(x) from the velocity function by taking the derivative:

a(x)=v'(x)

We can view the function

$v(x)=\int_{0}^{x^2+2x+1} z^d\sin(z^5) dz$

as the composition of the following functions

$g(x)=\int_{0}^{x} z^d\sin(z^5) dz$

h(x)=x^2+2x+1=(x+1)^2

so that v(x)=g(h(x)) . This means we use the chain rule

[g(h(x))]'=g'(h(x))h'(x)

to find the derivative. We have $g'(x)=x^d\sin x^5$ and h'(x)=2x+2 , so we have

$a(x)=v'(x)=[(x+1)^2]^d\sin [(x+1)^2]^5\cdot (2x+2)$

$=(2x+2)(x+1)^{2d}\sin (x+1)^{10}$

Report an Error

Example Question #2 : Calculate Position, Velocity, And Acceleration

The position of an object is given by the equation $y = \frac{1}{3} t^3 - \frac{1}{2} t^2 -2t + 4$ . What is its acceleration at t=2 ?

Possible Answers:

Correct answer:

Explanation:

If this function gives the position, the first derivative will give its speed and the second derivative will give its acceleration.

y ' = t^2 - t -2

y'' = 2t - 1

Now plug in 2 for t:

y'' = 2(2) - 1 = 4- 1 = 3

Report an Error

Example Question #1 : Calculate Position, Velocity, And Acceleration

The equation $y= \sin (2t)$ models the position of an object after t seconds. What is its speed after $\pi$ seconds?

Possible Answers:

Correct answer:

Explanation:

If this function gives the position, the first derivative will give its speed.

$y' = \cos (2t ) \cdot 2$

Plug in $\pi$ for t:

$2 \cos (2 \pi ) = 2\cdot 1 = 2$

Report an Error

Example Question #3 : Calculate Position, Velocity, And Acceleration

The position of an object is modeled by the equation $y= \frac{1}{2} \cos ^2 t$ What is the speed after $\frac{\pi}{4}$ seconds?

Possible Answers:

$-\frac{1}{2}$

$-\frac{1}{\sqrt2}$

$\sqrt{2}$

Correct answer:

$-\frac{1}{2}$

Explanation:

If this function gives the position, the first derivative will give its speed. To differentiate, use the chain rule: $f(g(x)) ' = f'(g(x))\cdot g'(x)$ . In this case, $f(x) = \frac{1}{2} x^2$ and $g(x) = \cos x$ . Since f'(x) = x and $g'(x ) = - \sin (x)$ , the first derivative is $\cos t \cdot - \sin t$ .

Plug in $\frac{\pi}{4}$ for t:

$\cos (\frac{\pi}{4} ) \cdot - \sin (\frac{\pi}{4} ) = \frac{1}{\sqrt{2}} \cdot - \frac{1}{\sqrt2} = -\frac{1}{2}$

Report an Error

Example Question #1 : Calculate Position, Velocity, And Acceleration

A particle's position on the -axis is given by the function $f(t)=\sin(t)-t+1$ from $0 < t <\pi$ .

When does the particle change direction?

Possible Answers:

$x=2\pi$

$x= \frac{\pi}{4}$

$x= \pi$

$x=\frac{\pi}{2}$

It doesn't change direction within the given bounds

Correct answer:

It doesn't change direction within the given bounds

Explanation:

To find when the particle changes direction, we need to find the critical values of f(t) . This is done by finding the velocity function, setting it equal to , and solving for

$0=v(t) = f'(t) = \cos(t)-1$ .

Hence $\cos(t)=1$ .

The solutions to this on the unit circle are $t =0, \pi$ , so these are the values of where the particle would normally change direction. However, our given interval is $0 < t < \pi$ , which does not contain $0, \pi$ . Hence the particle does not change direction on the given interval.

Report an Error

Example Question #4 : Calculate Position, Velocity, And Acceleration

A particle moves in space with velocity given by

$v(t)=\alpha t^3+\beta \sqrt{t}$

where $\alpha, \beta$ are constant parameters.

Find the acceleration of the particle when t=4 .

Possible Answers:

$48\alpha +2\beta$

$3\alpha t^2+\frac{\beta}{2\sqrt{t}}$

$48\alpha + \frac{\beta}{4}$

Correct answer:

$48\alpha + \frac{\beta}{4}$

Explanation:

To find the acceleration of the particle, we must take the first derivative of the velocity function:

$a(t)=v'(t)=3\alpha t^2+\frac{\beta}{2\sqrt{t}}$

The derivative was found using the following rule:

$\frac{\mathrm{d} }{\mathrm{d} x} x^n=nx^{n-1}$

Now, we evaluate the acceleration function at the given point:

$a(4)=3\alpha(4^2)+\frac{\beta}{2\sqrt{4}}=48\alpha+\frac{\beta}{4}$

Report an Error

View Tutors

PV
Certified Tutor

Abilene Christian University, Bachelor of Science, Engineering Physics.

View Tutors

Heather
Certified Tutor

University of West Alabama, Bachelor in Arts, English. The University of Alabama, Master of Science, English.

View Tutors

Nimil
Certified Tutor

University of Calgary, Master's/Graduate, Mechanical and Manufacturing Engineering.

All Calculus AB Resources

45 Practice Tests Question of the Day Flashcards Learn by Concept

Popular Subjects

SAT Tutors in Philadelphia, Math Tutors in Los Angeles, Biology Tutors in Seattle, MCAT Tutors in New York City, Algebra Tutors in Atlanta, Physics Tutors in Denver, Calculus Tutors in San Diego, ISEE Tutors in San Francisco-Bay Area, Spanish Tutors in Atlanta, MCAT Tutors in Seattle

Popular Courses & Classes

Spanish Courses & Classes in Seattle, SSAT Courses & Classes in Phoenix, Spanish Courses & Classes in Denver, SAT Courses & Classes in Phoenix, SSAT Courses & Classes in Denver, ISEE Courses & Classes in Miami, GMAT Courses & Classes in San Francisco-Bay Area, Spanish Courses & Classes in San Francisco-Bay Area, LSAT Courses & Classes in Miami, SAT Courses & Classes in Los Angeles

Popular Test Prep

MCAT Test Prep in New York City, GMAT Test Prep in Los Angeles, MCAT Test Prep in Phoenix, GRE Test Prep in Chicago, ACT Test Prep in New York City, ISEE Test Prep in San Francisco-Bay Area, LSAT Test Prep in Miami, ACT Test Prep in Phoenix, GMAT Test Prep in San Francisco-Bay Area, MCAT Test Prep in Seattle

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

Calculus AB : Calculate Position, Velocity, and Acceleration

Study concepts, example questions & explanations for Calculus AB

All Calculus AB Resources

Example Questions

Example Question #1 : Calculate Position, Velocity, And Acceleration

Example Question #1 : Contextual Applications Of Derivatives

Example Question #1 : Calculate Position, Velocity, And Acceleration

Example Question #1 : Calculate Position, Velocity, And Acceleration

Example Question #2 : Calculate Position, Velocity, And Acceleration

Example Question #2 : Calculate Position, Velocity, And Acceleration

Example Question #1 : Calculate Position, Velocity, And Acceleration

Example Question #3 : Calculate Position, Velocity, And Acceleration

Example Question #1 : Calculate Position, Velocity, And Acceleration

Example Question #4 : Calculate Position, Velocity, And Acceleration

All Calculus AB Resources

Report an issue with this question

DMCA Complaint

Contact Information

Address

Complaint Details

Email address:
Your name:
Feedback: