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

← Previous 1 2 3 4 5 6 7 8 9 … 34 35 Next →

Example Question #41 : Acceleration

The displacement of an object at time is defined by the equation $f(t) = \frac{100}{t}$ . What is the acceleration equation for this object?

Possible Answers:

$\frac{100}{t^3}$

$\frac{-1}{t^3}$

$\frac{-100}{t^3}$

$\frac{200}{t^3}$

Correct answer:

$\frac{200}{t^3}$

Explanation:

The acceleration equation is the second derivative of the displacement equation.

$f(t) = \frac{100}{t} = 100t^{-1}$

Therefore the first derivative is equal to

$f'(t) = (-1)(100)t^{-1-1} = -100t^{-2}$

Differentiating a second time gives

$f''(t) = (-2)(-100)t^{-2-1} = 200t^{-3} = \frac{200}{t^3}$

Report an Error

Example Question #42 : Acceleration

What is the acceleration of an object when its velocity is constantly increasing at a rate of two millimeters per second?

Possible Answers:

$\infty$

$\frac{1}{2}$

Correct answer:

Explanation:

When an object is constantly increasing at a certain rate, it is accelerating at that rate.

Write the equation that represents the object's velocity, where is in seconds.

v(t)=2t

To determine the acceleration, take the derivative of the velocity.

a(t)=2

The acceleration is 2 when the object is constantly increasing at 2 millimeters per second.

Report an Error

Example Question #41 : How To Find Acceleration

The position equation of an object is $f(t) = \sin(t)\cos(t)$ . What is the acceleration of the object at $t=\frac{\pi}{2}$ ?

Possible Answers:

$f''\left(\frac{\pi}{2}\right) = 1$

$f''\left(\frac{\pi}{2}\right) = -4$

$f''\left(\frac{\pi}{2}\right) = 2$

$f''\left(\frac{\pi}{2}\right) = 0$

Correct answer:

$f''\left(\frac{\pi}{2}\right) = 0$

Explanation:

The acceleration of the object is the second derivative of the position equation. To take the derivative of this function, we must use the product rule.

The product rule says

$f(x) = h(x)g(x) \Rightarrow f'(x) = h'(x)g(x) + h(x)g'(x)$ .

From the position equation $f(t) = \sin(t)\cos(t)$ , the first derivative is

$f'(t) = \cos(t)\cos(t) + \sin(t)(-\sin(t)) = \cos^2(t)-\sin^2(t)$ .

To find the second derivative, we must use the chain rule and the power rule.

The chain rule says $f(x) = h(g(x)) \rightarrow f'(x) = h'(g(x))\times g'(x)$ .

The power rule says that $f(x) = x^n \rightarrow f'(x) = nx^{n-1}$ .

Applying these rules to f'(t) gives,

$f''(t) = 2\cos(t)(-\sin(t)) - 2\sin(t)\cos(t) \Rightarrow \\ \Rightarrow -2\cos(t)\sin(t)-2\sin(t)\cos(t) = -4\sin(t)\cos(t)$

We now use $\frac {\pi}{2}$ as our value for to find our answer.

$f''\left(\frac{\pi}{2}\right) = -4\sin\left(\frac{\pi}{2}\right)\cos\left(\frac{\pi}{2}\right) = -4(1)(0) = 0$

Report an Error

Example Question #41 : How To Find Acceleration

The velocity of an object is given by the equation $v(t) = 7\cos(t)$ . What is the acceleration of the object?

Possible Answers:

$v'(t) = -7\sin(t)$

$v'(t) = 7\cos(t)\sin(t)$

$v'(t) = -7\cos(t)$

$v'(t) = -7\cos(t)\sin(t)$

Correct answer:

$v'(t) = -7\sin(t)$

Explanation:

The acceleration of the object can be found by taking the derivative of the velocity equation.

Using the velocity equation $v(t) = 7\cos(t)$ , the acceleration equation is:

$v'(t) = -7\sin(t)$ .

Report an Error

Example Question #45 : Acceleration

The velocity of an object is given by the equation v(t) = 4t-3 . What is the acceleration of the object at t = 5 ?

Possible Answers:

Correct answer:

Explanation:

Given the velocity equation v(t) = 4t-3 , the acceleration can be found by differentiating the velocity. We can do this using the power rule, where if

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

This can be done only for terms in the equation that have the variable . All other terms become .

Therefore,

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

Therefore, the acceleration at t = 5 is,

a(5) = 4 .

Report an Error

Example Question #46 : Acceleration

The position of an object is given by the equation x(t) = 100t^3-1000t^2 . What is the acceleration of the object at t = 10 ?

Possible Answers:

12000

700

4000

5000

Correct answer:

4000

Explanation:

Given the position equation x(t) = 100t^3-1000t^2 , the acceleration of the object can be found by differentiating twice. This can be done by using the power rule where if

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

Therefore, the first derivative is

$v(t) = 100(3)t^{(3-1)}- 1000(2)t^{(2-1)}= 300t^2 - 2000t$ .

Differentiating again gives us the acceleration equation.

$a(t) = 300(2)t^{(2-1)} - 2000(1)t^{(1-1)}= 600t - 2000$

We can now set time t= 10 to find the acceleration.

a(10) = 600(10) - 2000 = 6000 - 2000 = 4000

Report an Error

Example Question #47 : Acceleration

The velocity of an object is given by the equation v(t) = 4t^2 - 2 . 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 velocity equation v(t) = 4t^2 -2 .

To differentiate this equation we must use the power rule where

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

This rule applies to the first term in the velocity equation. The other term does not have the variable and therefore becomes when differentiated.

The derivative of the velocity equation is then,

$a(t) =v'(t) = 4(2)t^{2-1} - 0 = 8t$ .

We now use the value of for to find the acceleration at t = 2 .

a(2) = 8(2) = 16

Report an Error

Example Question #48 : Acceleration

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

Possible Answers:

$a(t) = 3t - \sin(t)$

$a(t) = 6t + \sin(t)$

$a(t) = 3 - \sin(t)$

$a(t) = 6t - \sin(t)$

Correct answer:

$a(t) = 6t + \sin(t)$

Explanation:

The acceleration equation can be found by differentiating the position equation twice. The first derivative will be the velocity equation, and the second derivative will be the acceleration equation.

To differentiate the position equation we must use the power rule where if

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

We must also use the fact that when the function $\sin(t)$ is differentiated, it becomes $\cos(t)$ , and when $\cos(t)$ is differentiated it becomes $-\sin(t)$ .

Applying these rules to the position equation gives us,

$x'(t) = v(t) = (3)t^{(3-1)} - (\cos(t)) = 3t^2 - \cos(t)$ .

Differentiating a second time gives,

$a(t) = v'(t) = 3(2)t^{(2-1)} - (-\sin(t)) = 6t + \sin(t)$ .

Report an Error

Example Question #49 : Acceleration

The jerk of an object is given by the equation j(x) = 10t -100 . What is the acceleration of the object at t = 2 , if the initial acceleration is ?

Possible Answers:

Correct answer:

Explanation:

The acceleration of the object can be found by integrating the jerk equation j(t) = a'(t) = 10t -100 . The jerk equation can be rewritten as j(t) = 10t^1 - 100t^0 . To integrate the equation we must use the power rule where

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

Applying this to the jerk equation gives us,

$a(t) = \int(10t^1 - 100t^0)dt= \frac{10t^{1+1}}{1+1} - \frac{100t^{0+1}}{0+1} + C$

$= \frac{10}{2}t^2 - \frac{100}{1}t^1+C \rightarrow$

$\rightarrow 5t^2 - 100t + C$ .

To complete the equation, we must find the value of using the initial acceleration.

a(0) = 5(0)^2 - 100(0) + C = 5

Therefore C = 5 and the acceleration equation is

a(t) = 5t^2 -100t +5 .

We can now find the acceleration at t = 2 .

a(2) = 5(2)^2 - 100(2) +5 = 20 - 200 +5 = -175

Report an Error

Example Question #50 : Acceleration

The position of an object is given by the equation $x(t) = \sin(2t) - e^{4t}$ . What is the acceleration of the object?

Possible Answers:

$a(t) = -4\sin(2t) - 16e^{4t}$

$a(t) = 4\sin(2t) + 16e^{4t}$

$a(t) = -\sin(2t) - e^{4t}$

$a(t) = 4\sin(2t) - 16e^{4t}$

Correct answer:

$a(t) = -4\sin(2t) - 16e^{4t}$

Explanation:

The acceleration of the object can be found by differentiating the position equation $x(t) = \sin(2t) - e^{4t}$ twice. To differentiate the position equation, we must use the chain rule where if

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

To differentiate the position equation we must also know that the derivative of $\sin(t)$ is $\cos(t)$ , the derivative of $\cos(t)$ is $-\sin(t)$ , and the derivative of e^t is e^t . Applying this to each term of the position equation gives us,

$x'(t) = \cos(2t)(2) - e^{4t}(4) = 2\cos(2t) - 4e^{4t}$ .

Differentiating the derivative of the position equation will give us the acceleration equation. To do this we must again use the chain rule.

$a(t) = x''(t) = 2(-\sin(2t))(2) - 4e^{4t}(4) = -4\sin(2t) - 16e^{4t}$

Therefore, the acceleration equation is

$a(t) = -4\sin(2t) - 16e^{4t}$ .

Report an Error

← Previous 1 2 3 4 5 6 7 8 9 … 34 35 Next →

View Calculus Tutors

Dawn
Certified Tutor

University of Oregon, Bachelor in Arts, Biochemistry and Molecular Biology. Johns Hopkins University, Doctor of Philosophy, E...

View Calculus Tutors

Emily
Certified Tutor

University of Maryland-College Park, Bachelor of Science, Education.

View Calculus Tutors

Sree
Certified Tutor

Saint Louis University-Main Campus, Bachelor in Arts, Chemistry.

All Calculus 1 Resources

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

Popular Subjects

Math Tutors in Denver, ISEE Tutors in Seattle, Spanish Tutors in San Diego, Physics Tutors in Los Angeles, Statistics Tutors in Miami, SAT Tutors in Boston, French Tutors in Chicago, Chemistry Tutors in Seattle, Algebra Tutors in Washington DC, Math Tutors in San Diego

Popular Courses & Classes

GMAT Courses & Classes in Washington DC, GRE Courses & Classes in Denver, MCAT Courses & Classes in Phoenix, GRE Courses & Classes in Seattle, GRE Courses & Classes in Boston, ISEE Courses & Classes in Miami, GMAT Courses & Classes in San Diego, SAT Courses & Classes in San Diego, GMAT Courses & Classes in Houston, ACT Courses & Classes in Boston

Popular Test Prep

ISEE Test Prep in San Francisco-Bay Area, ACT Test Prep in Boston, SSAT Test Prep in Los Angeles, LSAT Test Prep in Philadelphia, GMAT Test Prep in Houston, SSAT Test Prep in Seattle, LSAT Test Prep in New York City, SAT Test Prep in New York City, ISEE Test Prep in Seattle, GRE 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 #41 : Acceleration

Example Question #42 : Acceleration

Example Question #41 : How To Find Acceleration

Example Question #41 : How To Find Acceleration

Example Question #45 : Acceleration

Example Question #46 : Acceleration

Example Question #47 : Acceleration

Example Question #48 : Acceleration

Example Question #49 : Acceleration

Example Question #50 : Acceleration

All Calculus 1 Resources

Report an issue with this question

DMCA Complaint

Contact Information

Address

Complaint Details

Email address:
Your name:
Feedback: