Call Now to Set Up Tutoring:

(888) 888-0446

Calculus 1 : Calculus

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 #80 : 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) = 6t

a(t) = 0

a(t) = -6

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

Example Question #81 : How To Find Acceleration

The position of an object is given by the equation $x(t) = e^{(2t)}\sin(3t)$ . What is the acceleration of the object?

Possible Answers:

$a(t) = 10e^{(2t)}\sin(3t)+6e^{(2t)}\cos(3t)-9e^{(2t)}\sin(3t)$

$a(t) =12e^{(2t)}\cos(3t)-5e^{(2t)}\sin(3t)$

$a(t) = 4e^{(2t)}\sin(3t)+6e^{(2t)}\cos(3t)-9e^{(2t)}\sin(3t)$

$a(t) = 4e^{(2t)}\sin(3t)+6e^{(2t)}\cos(3t)+3e^{(2t)}\sin(3t)$

Correct answer:

$a(t) =12e^{(2t)}\cos(3t)-5e^{(2t)}\sin(3t)$

Explanation:

The acceleration of the object can be found by differentiating the position equation twice. This can be done using the power rule, chain rule, and product rule.

The power rule is

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

The chain rule is

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

The product rule is

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

Using these three rules we can find the velocity equation from the position equation.

$v(t) = e^{(2t)}(2t^{(1-1)})\sin(3t) + e^{(2t)}\cos(3t)(3t^{(1-1)}) = 2e^{(2t)}\sin(3t) +3e^{(2t)}\cos(3t)$

Repeating this process gives us the acceleration

$\\a(t) = 2e^{(2t)}(2t^{(1-1)})\sin(3t) + 2e^{(2t)}\cos(3t)(3t^{(1-1)}) +\\ \\ 3e^{(2t)}(2t^{(1-1)})\cos(3t)+3e^{(2t)}(-\sin(3t))(3t^{(1-1)}) =\\ \\ 4e^{(2t)}\sin(3t)+12e^{(2t)}\cos(3t)-9e^{(2t)}\sin(3t)\\ \\ =12e^{2t}\cos(3t)-5e^{2t}\sin(3t)$

Report an Error

Example Question #82 : How To Find Acceleration

If the acceleration function of an object is described by $a(t)=cos(2\pi)t$ , what is the acceleration when t=2 ?

Possible Answers:

Correct answer:

Explanation:

No deriving or integrating is necessary for this question.

Substitute the value of time into $a(t)=cos(2\pi)t$ .

$a(t=2)=cos(2\pi)(2)= 2cos(2\pi)= 2(1)=2$

Report an Error

Example Question #471 : Calculus

Find the acceleration function of an object if the position function is x(t)= -ln(t)+cos(2t) .

Possible Answers:

$\frac{1}{t^2}-\frac{1}{4}cos(2t)$

$-\frac{1}{t^2}+\frac{1}{4}cos(2t)$

$\frac{1}{t^2}+4cos(2t)$

$\frac{1}{t^2}-4cos(2t)$

$-\frac{1}{t^2}-4cos(2t)$

Correct answer:

$\frac{1}{t^2}-4cos(2t)$

Explanation:

To find the acceleration function from the position curve, take the derivative twice.

$v(t)= \frac{d}{dx} x(t)$

x(t)= -ln(t)+cos(2t)

Recall the derivative of the natural log is a special case.

$f(x)=ln(u)\rightarrow f'(x)=\frac{1}{u}\cdot \frac{du}{dx}$

Therefore, the velocity functions becomes:

$v(t)=\frac{d}{dx}x(t)= -\frac{1}{t}-2sin(2t)= -t^-^1-2sin(2t)$

$a(t)=t^-^2-4cos(2t) =\frac{1}{t^2}-4cos(2t)$

Report an Error

Example Question #81 : Acceleration

The veolcity of an object is given by the equation v(t) = 28t^3 + 10t^2 . What is the acceleration equation of the object?

Possible Answers:

$a(t) = 7t^4 + \frac{10t^3}{3}$

a(t) = 84t^2 + 20t

a(t) = 168t + 20

a(t) = 28t^4 + 10t^3

Correct answer:

a(t) = 84t^2 + 20t

Explanation:

The acceleration can be found by differentiating the velocity equation using the power rule where if

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

Using this rule we find the acceleration to be

$a(t) = 28(3)t^{(3-1)} + 10(2)t^{(2-1)} = 84t^2 + 20t$ .

Report an Error

Example Question #85 : How To Find Acceleration

The position of an object is given by the function $x(t) = \sin(t)\cos(t)$ . What is the acceleration of the object at t =1 ?

Possible Answers:

$-2\sin(1)\cos^2(1) + 2\sin^2(1)\cos(1)$

$-4\sin(1)\cos(1)$

$-2\sin(1)\cos^2(1) - 2\sin^2(1)\cos(1)$

$4\sin(1)\cos(1)$

Correct answer:

$-4\sin(1)\cos(1)$

Explanation:

The acceleration of the object can be found by differentiating the position twice using the product rule where if

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

Applying this rule to the position gives us the velocity.

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

To differentiate a second time we must 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

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

Therefore the acceleration of the object is

$\\a(t) = (2)\cos^{2-1}(t)(-\sin(t)) - (2)\sin^{2-1}(t)(\cos(t)) \Rightarrow \\ \\ -2\cos(t)\sin(t) - 2\sin(t)\cos(t) = -4\sin(t)\cos(t)$ .

The acceleration at t =1 is then

$a(t) = -4\sin(1)\cos(1)$ .

Report an Error

Example Question #86 : How To Find Acceleration

The position of an object is given by the equation x(t) = 10t^3 - t^4 . What is the acceleration of the object at t =1 ?

Possible Answers:

Correct answer:

Explanation:

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

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

Using this rule we find the velocity equation to be

$v(t) = 10(3)t^{(3-1)} - (4)t^{(4-1)} = 30t^2 - 4t^3$ .

Repeating this process gives the acceleration equation

$a(t) = 30(2)t^{(2-1)} - 4(3)t^{(3-1)} = 60t - 12t^2$ .

Therefore the acceleration at t =1 is

a(1) = 60(1) - 12(1)^2 = 60 - 12 = 48 .

Report an Error

Example Question #87 : How To Find Acceleration

The position of an object is given by the equation x(t) = 20t^2 - 4t^3 + 5 . 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 of the object twice using the power rule, where if

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

Applying this to the position equation gives us

$v(t) = 20(2)t^{2-1} - 4(3)t^{3-1} + 5(0)t^{(0-1)}= 40t - 12t^2$ .

Repeating this process gives us the acceleration equation

$a(t) = 40(1)t^{(1-1)} - 12(2)t^{(2-1)} = 40 - 24t$ .

Therefore the acceleration at t =2 is

a(2) = 40 - 24(2) = 40 -48 = -8 .

Report an Error

Example Question #88 : How To Find Acceleration

The position of an object is given by the equation x(t) = 10t - 2t^2 - 5 . What is the acceleration of the object at t = 4 ?

Possible Answers:

Correct answer:

Explanation:

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

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

This rule gives a velocity equation of

$v(t) = 10(1)t^{(1-1)} - 2(2)t^{(2-1)} - 5(0)t^{(0-1)} = 10 - 4t$ .

Repeating this process give us an acceleration of

$a(t) = 10(0)t^{(0-1)} - 4(1)t^{(1-1)} = -4$ .

Therefore the acceleration of the object is constant at .

Report an Error

Example Question #89 : How To Find Acceleration

The position of an object is given by the equation x(t) = 4t- 20 . What is the acceleration of the object at t =1 ?

Possible Answers:

Correct answer:

Explanation:

The acceleration of the object can be by differentiating the position equation using the power rule where if

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

Applying the power rule to the position equation gives

$v(t) = 4(1)t^{(1-1)} - 20(0)t^{(0-1)} = 4 - 0 = 4$ .

Repeating this process gives us the acceleration equation

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

Therefore the acceleration of the object is a constant .

Report an Error

View Calculus Tutors

Ping
Certified Tutor

The University of Texas at Dallas, Bachelor of Science, Computer Science. University of North Texas, Master of Arts, Education.

View Calculus Tutors

Ezam
Certified Tutor

University of the West Indies, Bachelor of Science, Mathematics and Computer Science. University of Phoenix-Central Valley Ca...

View Calculus Tutors

Rodolfo
Certified Tutor

The University of Texas at San Antonio, Bachelor of Science, Mathematics.

All Calculus 1 Resources

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

Popular Subjects

Spanish Tutors in Miami, French Tutors in Houston, Biology Tutors in Los Angeles, English Tutors in Chicago, SAT Tutors in Dallas Fort Worth, French Tutors in New York City, MCAT Tutors in New York City, Math Tutors in San Diego, English Tutors in Los Angeles, Chemistry Tutors in New York City

Popular Courses & Classes

Spanish Courses & Classes in Philadelphia, MCAT Courses & Classes in Denver, GMAT Courses & Classes in San Francisco-Bay Area, Spanish Courses & Classes in Miami, LSAT Courses & Classes in Denver, MCAT Courses & Classes in Chicago, GRE Courses & Classes in Miami, SSAT Courses & Classes in Boston, ISEE Courses & Classes in Seattle, ACT Courses & Classes in Chicago

Popular Test Prep

MCAT Test Prep in Dallas Fort Worth, MCAT Test Prep in San Francisco-Bay Area, SAT Test Prep in Phoenix, GRE Test Prep in Boston, ACT Test Prep in Chicago, GMAT Test Prep in Philadelphia, GRE Test Prep in Philadelphia, SSAT Test Prep in Seattle, GRE Test Prep in Atlanta, GRE Test Prep in Los Angeles

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 : Calculus

Study concepts, example questions & explanations for Calculus 1

All Calculus 1 Resources

Example Questions

Example Question #80 : How To Find Acceleration

Example Question #81 : How To Find Acceleration

Example Question #82 : How To Find Acceleration

Example Question #471 : Calculus

Example Question #81 : Acceleration

Example Question #85 : How To Find Acceleration

Example Question #86 : How To Find Acceleration

Example Question #87 : How To Find Acceleration

Example Question #88 : How To Find Acceleration

Example Question #89 : 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: