We are open Saturday and Sunday!

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 #301 : How To Find Acceleration

If $x(t)=\frac{1}{4}x^2-2x+1$ , what is the acceleration function?

Possible Answers:

$a(t)=\frac{1}{2}x$

a(t)=3

$a(t)=-\frac{1}{2}$

$a(t)=\frac{1}{2}$

a(t)=1

Correct answer:

$a(t)=\frac{1}{2}$

Explanation:

To find the acceleration, you must first find the velocity function, which is the derivative of position. To take the derivative, multiply the exponent by the coefficient in front of the x and then decrease the exponent by 1. Therefore, the velocity function is: $v(t)=\frac{1}{2}x-2$ . To find acceleration now, take the derivative of the velocity function, which is $a(t)=\frac{1}{2}$ .

Report an Error

Example Question #302 : How To Find Acceleration

Find the acceleration of a particle whose position (meters) is described by s(t)=6t^2+4t+1. , and time is in seconds.

Possible Answers:

$24 \frac{m}{s^{2}}$

$30\frac{m}{s^{2}}$

$(12t+4) \frac{m}{s^{2}}$

$12 \frac{m}{s^{2}}$

$16 \frac{m}{s^{2}}$

Correct answer:

$12 \frac{m}{s^{2}}$

Explanation:

The acceleration is the second derivative of the position. The derivative of a number (without any variables) is zero. Therefore, all of the terms drop out except for the leading term.

$s'(t)=12t+4. \;s''(t)=12\frac{m}{s^{2}}.$

Report an Error

Example Question #303 : How To Find Acceleration

The position of a particle is described by s(t)=(4+5t)^2 . Find the velocity and the acceleration as a function of time.

Possible Answers:

v(t)=30t+16, a(t)=30

v(t) = 4+5t, a(t)=5

v(t)=25t+16, a(t)=25

v(t)=50t, a(t)=0

v(t)=50t+40, a(t)=50

Correct answer:

v(t)=50t+40, a(t)=50

Explanation:

The velocity can be found by taking the first derivative of the position function. The second derivative determines the acceleration. This derivative involves the chain rule. s'(t)=2(4+5t)*5=10(4+5t)=40+50t .

s''(t)=a(t)=50.

Report an Error

Example Question #301 : Acceleration

The position of t= 1 is given by the following function:

$p(t)= (3x^2 + e^{2x})^2$

Find the acceleration.

Possible Answers:

1602.3

124.7

4319.4

Answer not listed

234.2

Correct answer:

1602.3

Explanation:

In order to find the acceleration of a certain point, you first find the derivative of the position function to get the velocity function; and then the derivative of the velocity function to find the acceleration function: p''(t)=v'(t)=a(t)

In this case, the position function is: $p(t)= (3t^2 + e^{2t})^2$

Then take the derivative of the position function to get the velocity function: $v(t)= 2(3t^2 + e^{2t}) (6t+2e^{2t})$

Then take the derivative of the velocity function to get the acceleration function: $a(t)= 2(6t+2e^{2t})^2 + 2(3t^2 + e^{2t}) (4e^{2t}+6)$

Then, plug t= 1 into the acceleration function: $a(1)= 2(6({\color{Blue} 1})+2e^{2({\color{Blue} 1})})^2 + 2(3({\color{Blue} 1})^2 + e^{2({\color{Blue} 1})}) (4e^{2({\color{Blue} 1})}+6)$

Therefore, the answer is: 1602.3

Report an Error

Example Question #302 : How To Find Acceleration

The position of t= 0 is given by the following function:

$p(t)= \sin(t^2+3t^4)$

Find the acceleration.

Possible Answers:

1602.3

Answer not listed

21.5

Correct answer:

Explanation:

In this case, the position function is: $p(t)= \sin(t^2+3t^4)$

Then take the derivative of the position function to get the velocity function:

$v(t)= \cos(t^2+3t^4) (2t+12t^3)$

Then take the derivative of the velocity function to get the acceleration function:

$a(t)= (36t^2+2) \cos(3t^4+t^2) - (12t^3+2t)^2 \sin(3t^4+t^2)$

Then, plug t= 0 into the acceleration function: $a(0)= (36({\color{Blue} 0})^2+2) \cos(3({\color{Blue} 0})^4+ {\color{Blue} 0}^2) - (12({\color{Blue} 0})^3+2({\color{Blue} 0}))^2 \sin(3({\color{Blue} 0})^4+{\color{Blue} 0}^2)$

Therefore, the answer is:

Report an Error

Example Question #303 : How To Find Acceleration

The position of t= 0 is given by the following function:

$p(t)= 2(\frac{1}{2}t+3)^2$

Find the acceleration.

Possible Answers:

Answer not listed

Correct answer:

Explanation:

In this case, the position function is: $p(t)= 2(\frac{1}{2}t+3)^2$

Then take the derivative of the position function to get the velocity function: v(t)= t+6

Then take the derivative of the velocity function to get the acceleration function: a(t)= 1

Then, plug t= 0 into the acceleration function: a(0)= 1

Therefore, the answer is:

Report an Error

Example Question #307 : How To Find Acceleration

The position of t= 1 is given by the following function:

p(t)= -ln(2t^2 + 5t)

Find the acceleration.

Possible Answers:

7.4

Answer not listed

-3.3

1.1

2.5

Correct answer:

1.1

Explanation:

In this case, the position function is: p(t)= -ln(2t^2 + 5t)

Then take the derivative of the position function to get the velocity function: $v(t)= - \frac{4t+5}{2t^2 + 5t}$

Then take the derivative of the velocity function to get the acceleration function: $a(t)= \frac{(4t+5)^2}{(2t^2 + 5t)^2} - \frac{4}{2t^2+5t}$

Then, plug t= 1 into the acceleration function: $a(1)= \frac{(4({\color{Blue} 1})+5)^2}{(2({\color{Blue} 1})^2 + 5({\color{Blue} 1}))^2} - \frac{4}{2({\color{Blue} 1})^2+5({\color{Blue} 1})}$

Therefore, the answer is: 1.1

Report an Error

Example Question #302 : Acceleration

The position of t= 0 is given by the following function:

$p(t)= 4t^3 - \cos(t^2) + e^{2t}$

Find the acceleration.

Possible Answers:

Answer not listed

Correct answer:

Explanation:

In this case, the position function is: $p(t)= 4t^3 - \cos(t^2) + e^{2t}$

Then take the derivative of the position function to get the velocity function: $v(t)= 12t^2 - 2t\cos(t^2) + 2e^{2t}$

Then take the derivative of the velocity function to get the acceleration function: $a(t)= 4t^2\sin(t^2) - 2\cos(t^2) + 4e^{2t} + 24t$

Then, plug t= 0 into the acceleration function: $a(0)= 4({\color{Blue} 0})^2\sin({\color{Blue} 0}^2) - 2\cos({\color{Blue} 0}^2) + 4e^{2({\color{Blue} 0})} + 24({\color{Blue} 0})$

Therefore, the answer is:

Report an Error

Example Question #692 : Spatial Calculus

The position of t= 0 is given by the following function:

$p(t)= \cos(2t +e^t)$

Find the acceleration.

Possible Answers:

-3.002

4.252

-1.003

Answer not listed

Correct answer:

-3.002

Explanation:

In this case, the position function is: $p(t)= \cos(2t +e^t)$

Then take the derivative of the position function to get the velocity function: $v(t)= -\sin(2t +e^t) (2+e^t)$

Then take the derivative of the velocity function to get the acceleration function: $a(t)= -e^t \sin(e^t+2t) - (e^t+2)^2 \cos(e^t+2t)$

Then, plug t= 0 into the acceleration function: $a(t)= -e^{\color{Blue} 0} \sin(e^{\color{Blue} 0}+2({\color{Blue} 0})) - (e^{\color{Blue} 0}+2)^2 \cos(e^{\color{Blue} 0}+2({\color{Blue} 0}))$

Therefore, the answer is: -3.002

Report an Error

Example Question #310 : How To Find Acceleration

Find the acceleration function of the falling rock if its position is given by the following function:

s(t)=5-20t-8t^2

Possible Answers:

-8t

-20-16t

-16t

Correct answer:

Explanation:

To find the acceleration function of the falling rock, we must take the second derivative of the position function for the rock:

v(t)=s'(t)=-20-16t

a(t)=v'(t)=-16

The derivatives were found using the following rule:

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

Report an Error

View Calculus Tutors

Chukwudi
Certified Tutor

University of Nigeria, Nsukka., Bachelor of Science, Biochemistry. African University of Science and Technology., Master of S...

View Calculus Tutors

Abigail
Certified Tutor

Morehead State University, Bachelor of Science, Mathematics. The University of Tennessee, Doctor of Science, Nuclear Engineer...

View Calculus Tutors

Lamar
Certified Tutor

Cleveland State University, Bachelors, Mathematics.

All Calculus 1 Resources

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

Popular Subjects

SSAT Tutors in Chicago, ISEE Tutors in Los Angeles, Statistics Tutors in Miami, ISEE Tutors in San Francisco-Bay Area, Calculus Tutors in Philadelphia, ISEE Tutors in Houston, LSAT Tutors in Miami, SAT Tutors in Boston, Algebra Tutors in San Francisco-Bay Area, Biology Tutors in New York City

Popular Courses & Classes

GRE Courses & Classes in San Francisco-Bay Area, ISEE Courses & Classes in Boston, GMAT Courses & Classes in New York City, SAT Courses & Classes in Phoenix, Spanish Courses & Classes in Atlanta, GRE Courses & Classes in Atlanta, SAT Courses & Classes in Seattle, MCAT Courses & Classes in Boston, MCAT Courses & Classes in Los Angeles, MCAT Courses & Classes in Washington DC

Popular Test Prep

SSAT Test Prep in New York City, MCAT Test Prep in Chicago, GRE Test Prep in San Francisco-Bay Area, SSAT Test Prep in Miami, ACT Test Prep in Dallas Fort Worth, GRE Test Prep in Dallas Fort Worth, MCAT Test Prep in Washington DC, LSAT Test Prep in New York City, ISEE Test Prep in Boston, GMAT 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

Calculus 1 : How to find acceleration

Study concepts, example questions & explanations for Calculus 1

All Calculus 1 Resources

Example Questions

Example Question #301 : How To Find Acceleration

Example Question #302 : How To Find Acceleration

Example Question #303 : How To Find Acceleration

Example Question #301 : Acceleration

Example Question #302 : How To Find Acceleration

Example Question #303 : How To Find Acceleration

Example Question #307 : How To Find Acceleration

Example Question #302 : Acceleration

Example Question #692 : Spatial Calculus

Example Question #310 : 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: