Call Now to Set Up Tutoring:

(888) 888-0446

Calculus 1 : How to find acceleration

Study concepts, example questions & explanations for Calculus 1

Create An Account

All Calculus 1 Resources

10 Diagnostic Tests 438 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

Example Question #71 : Acceleration

The position of an object is $x(t) = \sin(t)\cos(t)$ . What is the acceleration of the object?

Possible Answers:

a(t) = 0

$a(t) = -4\cos(t)\sin(t)$

None of the above.

$a(t) = 4\cos(t)\sin(t)$

Correct answer:

$a(t) = -4\cos(t)\sin(t)$

Explanation:

The acceleration of the object can be found by differentiating the position equation twice, using the product rule. The product rule is where

$f(x)= j(x)k(x) \rightarrow f'(x) = j'(x)k(x) + j(x)k'(x)$ .

Using the product rule, the velocity of the object is

$v(t) = (\cos(t))\cos(t) + \sin(t)(-\sin(t)) = \cos^2(t) - \sin^2(t)$ .

To differentiate the velocity equation we must use the power rule and the chain rule.

The power rule is

$f(x) = x^n \rightarrow f'(x) = nx^{n-1}$ .

The chain rule is

$f(x) = h(g(x)) \rightarrow f'(x) = h'(g(x) \times g'(x)$ .

Therefore the acceleration equation is then

$\\a(t) = v'(t) = (2)\cos^{(2-1)}(t)(-\sin(t)) - (2)\sin^{(2-1)}(t)(\cos(t)) \\ \Rightarrow -2\cos(t)\sin(t) - 2\sin(t)\cos(t) = -4\cos(t)\sin(t)$ .

Report an Error

Example Question #71 : How To Find Acceleration

The velocity of an object is given by the equation v(t) = 2t + 5 . What is the acceleration of the object at time t = 4 ?

Possible Answers:

None of these.

Correct answer:

None of these.

Explanation:

The acceleration of the object is the derivative of the velocity. We can differentiate the velocity using the power rule where if

$f(x) = x^n \rightarrow f'(x) = nx^{n-1}$ .

Therefore the acceleration equation of the object is

$a(t) = 2t^1 - 5t^0 =2(1)t^{(1-1)} - 5(0)t^{(0-1)} = 2$ .

Because the acceleration does not have depend on the variable the function is constant. Therefore the acceleration of the object is .

Report an Error

Example Question #72 : How To Find Acceleration

The position of an object is given by the function x(t) = 5t^5 - 3t^2 . What is the acceleration of the object at time t =1 ?

Possible Answers:

Correct answer:

Explanation:

The acceleration of the object can be found by differentiating the position equation of the object twice. This can be done using the power rule where if

$f(x) = x^n \rightarrow f'(x) = nx^{n-1}$ .

Therefore the velocity equation is

$v(t) = 5(5)t^{(5-1)} - 3(2)t^{(2-1)} = 25t^4 - 6t^$ .

Repeating this process gives the acceleration equation

$a(t) = 25(4)t^{(4-1)} - 6(1)t^{(1-1)} = 100t^3 - 6$ .

Now we can find the acceleration at t= 1 .

a(1) = 100(1)^3 - 6 = 100 -6 = 94

Report an Error

Example Question #73 : How To Find Acceleration

The jerk of an object is given by j(t) = 60t^3 . What is the acceleration of the object at time t= 2 , if the initial acceleration of the object is ?

Possible Answers:

249

200

Correct answer:

249

Explanation:

To find acceleration we integrate the jerk of the object. This can be done using the power rule where if

$f'(x) = x^n \rightarrow f(x) = \frac{x^{n+1}}{n+1}$ .

Therefore the acceleration is

$a(t) = \int (60t^3)dt = \frac{60t^{(3+1)}}{3+1} + C = 15t^4 +C$ .

The initial acceleration of the equation can now be used to find the value of .

a(0) = 15(0)^4 + C = 0 + C = 9

Therefore C = 9 and a(t) = 15t^4 + 9 .

The acceleration at time t =2 can now be found.

a(2) = 15(2)^4 + 9 = 249

Report an Error

Example Question #74 : How To Find Acceleration

The velocity of an object is $v(t) = e^{7t}$ . What is the acceleration of the object at t = 0 ?

Possible Answers:

e^7

Correct answer:

Explanation:

The acceleration can be found by differentiating the velocity.

The velocity can be differentiated using the chain rule where

$f(x) = h(g(x)) \rightarrow f'(x) = h'(g(x) \times g'(x)$ .

The acceleration is then

$a(t) = v'(t) = e^{7t}(7) = 7e^{7t}$ .

At t = 0 , the acceleration is

$a(0) = 7e^{7(0)} = 7e^0 = 7$ .

Report an Error

Example Question #75 : How To Find Acceleration

The position of an object is given by the equation x(t) = 10t -17 . What is the acceleration equation for the object?

Possible Answers:

a(t) = 10

None of these.

a(t) = 0

$a(t) = \frac{5t^3}{6} - \frac{17t^2}{2}$

Correct answer:

a(t) = 0

Explanation:

The acceleration can be found by differentiating the position equation twice using the power rule.

The power rule is where

$f(x) = x^n \rightarrow f'(x) = nx^{n-1}$ .

Therefore the velocity of the object is

$v(t) = x'(t) = 10(1)t^{(1-1)} - 17(0)t^{0-1} = 10$ .

The acceleration of the object can be found using the same method of differentiation as before.

$a(t) = v'(t) = 10(0)t^{(0-1)} = 0$

Therefore the acceleration of the object is .

Report an Error

Example Question #76 : How To Find Acceleration

The velocity of an object is $v(t) = \sin(e^{2t})$ . What is the acceleration of the object at time t = -1 ?

Possible Answers:

$\frac{2}{e^2}\cos(e^{-2})$

${2e^2}\cos(e^{2})$

$\frac{2}{e^2}\cos(e^{2})$

$2e^2\cos(2)$

Correct answer:

$\frac{2}{e^2}\cos(e^{-2})$

Explanation:

The acceleration of the object is the derivative of the velocity equation. To differentiate this equation we can use the chain rule and the power rule where if

$f(x) = h(g(x)) \rightarrow f'(x) = h'(g(x) \times g'(x)$

and if

$f(x) = x^n \rightarrow f'(x) = nx^{n-1}$ .

From these rules we find that the acceleration of the object is

$a(t) = \cos(e^{2t})(e^{2t})(2) = 2e^{2t}\cos(e^{2t})$ .

We can now find the acceleration at time t = -1 .

$a(-1) = 2e^{2(-1)}\cos(e^{2(-1)}) = \frac{2}{e^2}\cos(e^{-2})$

Report an Error

Example Question #77 : How To Find Acceleration

The position of an object is given by the equation x(t) = 4t^2 - t^3 . What is the acceleration of the object at t =2 ?

Possible Answers:

Correct answer:

Explanation:

The acceleration of the object can be found by differentiating the position twice. This can be done using the power rule where if

$f(x) = x^n \rightarrow f'(x) = nx^{n-1}$ .

Using this we can differentiate the velocity equation.

$v(t) = 4(2)t^{(2-1)} - (3)t^{(3-1)} = 8t - 3t^2$

Repeating this process we can find the acceleration equation.

$a(t) = 8(1)t^{(1-1)} - 3(2)t^{(2-1)} = 8 - 6t$

We can now find the acceleration at t = 2 .

a(2) = 8 - 6(2) = 8 - 12 = -4

Report an Error

Example Question #78 : How To Find Acceleration

The position of an object is given by the equation x(t) = 6t^3e^t . What is the acceleration of the object?

Possible Answers:

a(t) = 6te^t+36t^2e^t+6t^3e^t

a(t) = 72te^t+36t^2e^t+6t^3e^t

a(t) = 36te^t+18t^2e^t+6t^3e^t

a(t) = 36te^t+36t^2e^t+6t^3e^t

Correct answer:

a(t) = 36te^t+36t^2e^t+6t^3e^t

Explanation:

To find the acceleration of the object we must differentiate the position of the object twice. We can differentiate the position using the power rule and the product rule where if

$f(x) = x^n \rightarrow f'(x) = nx^{n-1}$

and if

$f(x)= j(x)k(x) \rightarrow f'(x) = j'(x)k(x) + j(x)k'(x)$ .

Using these rules we obtian

$v(t) = (6(3)t^{(3-1)})e^t + 6t^3(e^t) = 18t^2e^t+6t^3e^t$ .

Using the same rules we can differentiate again to give the acceleration equation.

$a(t) = 18(2)t^{(2-1)}e^t +18t^2(e^t) + 6(3)t^{(3-1)}e^t + 6t^3(e^t) = 36te^t+36t^2e^t+6t^3e^t$

Report an Error

Example Question #79 : How To Find Acceleration

The position of an object is given by the equation x(t) =3t^2 - 4t . What is the acceleration of the object?

Possible Answers:

a(t) = 6

a(t) = 6t

a(t) = 0

a(t) = -6

Correct answer:

a(t) = 6

Explanation:

The acceleration of the object can be found by differentiating the position equation twice. This can be done using the power rule where if

$f(x) = x^n \rightarrow f'(x) = nx^{n-1}$ .

Using this rule we find that

$v(t) = 3(2)t^{(2-1)} - 4(1)t^{(1-1)} = 6t - 4$ .

Repeating this process gives us

$a(t) = 6(1)t^{(1-1)} - 4(0)t^{(0-1)} = 6$ .

Report an Error

View Calculus Tutors

Andrew
Certified Tutor

University of Wisconsin-Stevens Point, Bachelor of Science, Mathematics. University of Wisconsin-Oshkosh, Master of Science, ...

View Calculus Tutors

Magdy
Certified Tutor

Cairo University, Egypt, Bachelor of Science, Electrical Engineering. New Mexico State University-Main Campus, Doctor of Phil...

View Calculus Tutors

Michael
Certified Tutor

Cleveland State University, Bachelor of Science, Mathematics.

All Calculus 1 Resources

10 Diagnostic Tests 438 Practice Tests Question of the Day Flashcards Learn by Concept

Popular Subjects

Computer Science Tutors in Los Angeles, Spanish Tutors in New York City, ACT Tutors in Phoenix, GRE Tutors in Philadelphia, ISEE Tutors in Chicago, Chemistry Tutors in Dallas Fort Worth, Computer Science Tutors in Denver, GRE Tutors in Atlanta, ISEE Tutors in Philadelphia, Algebra Tutors in Chicago

Popular Courses & Classes

SSAT Courses & Classes in Philadelphia, ACT Courses & Classes in Philadelphia, LSAT Courses & Classes in San Diego, SSAT Courses & Classes in Washington DC, SSAT Courses & Classes in Atlanta, ACT Courses & Classes in Phoenix, ACT Courses & Classes in Chicago, MCAT Courses & Classes in Atlanta, SAT Courses & Classes in Philadelphia, SAT Courses & Classes in New York City

Popular Test Prep

ACT Test Prep in Houston, SSAT Test Prep in Washington DC, SAT Test Prep in Phoenix, SSAT Test Prep in San Diego, SSAT Test Prep in Chicago, SAT Test Prep in Philadelphia, ACT Test Prep in Phoenix, LSAT Test Prep in Atlanta, SSAT Test Prep in Phoenix, 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 1 : How to find acceleration

Study concepts, example questions & explanations for Calculus 1

All Calculus 1 Resources

Example Questions

Example Question #71 : Acceleration

Example Question #71 : How To Find Acceleration

Example Question #72 : How To Find Acceleration

Example Question #73 : How To Find Acceleration

Example Question #74 : How To Find Acceleration

Example Question #75 : How To Find Acceleration

Example Question #76 : How To Find Acceleration

Example Question #77 : How To Find Acceleration

Example Question #78 : How To Find Acceleration

Example Question #79 : How To Find Acceleration

All Calculus 1 Resources

Report an issue with this question

DMCA Complaint

Contact Information

Address

Complaint Details

Email address:
Your name:
Feedback: