Call Now to Set Up Tutoring:

(888) 888-0446

Calculus 1 : 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 #291 : How To Find Acceleration

The position of a certain point is given by the following function:

$p(t)= 2t^{\frac{1}{2}} + 3e^t +4$

What is the acceleration at t=4 ?

Possible Answers:

163.7

170.2

342

165.8

160

Correct answer:

163.7

Explanation:

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

In this case, the position function is: $p(t)= 2t^{\frac{1}{2}} + 3e^t +4$

The velocity function is found by taking the derivative of the position function: $v(t)= t^{-\frac{1}{2}} + 3e^t$

The acceleration function is found by taking the derivative of the velocity function: $a(t)= {-\frac{1}{2}}t^{(-\frac{3}{2})} + 3e^t$

Finally, to find the accelaration, substitute t=4 into the acceleration function: $a(4)= {-\frac{1}{2}}(4)^{(-\frac{3}{2})} + 3e^{(4)}$

Therefore, the answer is: 163.7

Report an Error

Example Question #291 : Acceleration

The position of t=2 is given by the following function: p(t) = (t-4)^3

Find the acceleration.

Possible Answers:

Answer not listed

Correct answer:

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. Then find the derivative of the velocity function to get the acceleration function: p''(t)=v'(t) = a(t)

In this case, the position function is: p(t) = (t-4)^3

Then take the derivative of the position function to get the velocity function: v(t) = 3(t-4)^2

Then take the dervivative of the velocity function to get the acceleration function: a(t) = 6(t-4)

Then, plug t= 2 into the acceleration function: $a(2) = 6({\color{Blue} 2}-4)$

Therefore, the answer is:

Report an Error

Example Question #292 : Acceleration

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

p(t)= ln(2t)

Find the acceleration.

Possible Answers:

Answer not listed

Correct answer:

Explanation:

In this case, the position function is: p(t) = (t-4)^3

Then take the derivative of the position function to get the velocity function: $v(t)= \frac{1}{t}$

Then take the dervivative of the velocity function to get the acceleration function: $a(t)= -\frac{1}{t^2}$

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

Therefore, the answer is:

Report an Error

Example Question #683 : Spatial Calculus

A rocket begins at rest, and then begins to accelerate upwards at a constant rate. After 5 seconds, it has travelled 150 meters. How fast is it accelerating?

Possible Answers:

$12 \,m/s^2$

$60\,m/s^2$

$6\,m/s^2$

$30\,m/s^2$

Correct answer:

$12 \,m/s^2$

Explanation:

We'll need to first figure out how to express the rocket's position in terms of its acceleration, . This will give us an equation we can solve for .

The rocket begins at rest, so

$\begin{align*} p_0&=0\,m \\ v_0 &= 0\,m/s\end{align*}$

Initial and final time are given as

$\begin{align*} t_i &=0\,s \\ t_f &= 5\,s \end{align*}$

We're also given that

$p(5)=150\,m$

Acceleration is the rate of change in velocity, which is the rate of change in position, so to find the equation for position, we'll ned to integrate and use initial conditions twice.

$\int a(t)dt=\int a\,dt$

To integrate, we use the following rule:

$\int a\,dt = ax+C$

To solve for C, we'll use our initial condition, v(0)=0 .

$\\ v(0)=a(0)+C=0 \\ C=0 \\ v(t)=at$

We'll need to integrate again to find position:

$\int v(t)=\int at \, dt$

using the following rule:

$\int x^n\,dx=\frac{x^{n+1}}{n+1}+C$

Remember, when integrating we have to add the constant, C, to indicate that there's multiple functions that could be our answer.

Integrating then gives us:

$\begin{align*} \int v(t)\,dt &= \int at\,dt \\ &=\frac{at^2}{2}+C \\ &=\frac{a}{2}t^2+C \end{align*}$

To solve for C, we'll use our initial condition, p(0)=0 .

$\\ p(0)=\frac{a}{2}(0^2)+C=0 \\ C=0 \\ p(t)=\frac{a}{2}t^2$

Now, we know that p(5)=150 , so we can solve for .

$\begin{align*} \frac{a}{2}(5^2) &=150\\ 25a &= 300 \\ a &= 12\,m/s^2 \end{align*}$

Report an Error

Example Question #293 : Acceleration

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

$p(t)= e^{3t}$

Find the acceleration.

Possible Answers:

180.8

12.7

412.8

Answer not listed

327.5

Correct answer:

180.8

Explanation:

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

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

Then take the dervivative of the velocity function to get the acceleration function: $a(t)= 9e^{3t}$

Then, plug t= 1 into the acceleration function: $a(1)= 9e^{3({\color{Blue} 1})}$

Therefore, the answer is: 180.8

Report an Error

Example Question #294 : Acceleration

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

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

Find the acceleration.

Possible Answers:

Answer not listed

Correct answer:

Explanation:

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

Then take the derivative of the position function to get the velocity function: $v(t)=6t \cos(3t^2)$

Then take the dervivative of the velocity function to get the acceleration function: $a(t)=6 \cos(3t^2) - 36t^2\sin(3t^2)$

Then, plug t= 0 into the acceleration function: $a(0)=6 \cos(3({\color{Blue} 0})^2) - 36({\color{Blue} 0})^2\sin(3({\color{Blue} 0})^2)$

Therefore, the answer is:

Report an Error

Example Question #295 : Acceleration

If the position of a particle is given by $p(t)=\sin t \cos t$ , find the equation for its acceleration.

Possible Answers:

a(t)= 0

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

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

a(t)=-2

Correct answer:

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

Explanation:

Acceleration is the rate of change in velocity, which is the rate of change in position, so to answer this question, we'll need to differentiate this equation twice. First, we find the velocity, by applying the product rule

(fg)'=(f'g)+(fg')

This gives us:

$\begin{align*}v(t)&=(\sin t \cos t)'\\&= (\cos t \cdot \cos t)+(\sin t\cdot -\sin t)\\&= \cos^2 t-\sin^2 t\end{align*}$

Differentiating that using the chain rule

(f(g(t)))'=g'(t)f'(g(t))

on each term gives us

$\begin{align*} a(t)& =(\cos^2 t)'-(\sin^2 t)'\\ &=(-\sin t)(2\cos t)-(\cos t)(2\sin t)\\&=-4\sin t\cos t\end{align*}$

Report an Error

Example Question #296 : Acceleration

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

p(t)= 2t^4 + cos(t^2) - ln(2t)

Find the acceleration.

Possible Answers:

Answer not listed

45.9

25.5

12.4

27.4

Correct answer:

25.5

Explanation:

In this case, the position function is: p(t)= 2t^4 + cos(t^2) - ln(2t)

Then take the derivative of the position function to get the velocity function: $v(t)= 8t^3 - 2t\sin(t^2) - \frac{1}{t}$

Then take the dervivative of the velocity function to get the acceleration function: $a(t)= 24t^2 - 2\sin(t^2) + 4t^2\cos(t^2) + \frac{1}{t^2}$

Then, plug t= 1 into the acceleration function: $a(1)= 24({\color{Blue} 1})^2 - 2\sin({\color{Blue} 1}^2) + 4({\color{Blue} 1})^2\cos({\color{Blue} 1}^2) + \frac{1}{{\color{Blue} 1}^2}$

Therefore, the answer is: 25.5

Report an Error

Example Question #297 : Acceleration

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

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

Find the acceleration.

Possible Answers:

53.5

37.6

32.7

Answer not listed

72.8

Correct answer:

37.6

Explanation:

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

Then take the derivative of the position function to get the velocity function: $v(t)= 2e^{2t} - \frac{1}{t ln^2(2t)}$

Then take the dervivative of the velocity function to get the acceleration function: $a(t)= 4e^{2t} + \frac{1}{t^2 ln^2(2t)} + \frac{2}{t^2ln^3(2t)}$

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

Therefore, the answer is: 37.6

Report an Error

Example Question #298 : Acceleration

The position of a particle is given by the equation p(t)=t^3+3t^2-15t+6 . What is the particle's acceleration at t=1 ?

Possible Answers:

$6\,m/s^2$

$-6\,m/s^2$

$12\,m/s^2$

$-5\, m/s^2$

Correct answer:

$12\,m/s^2$

Explanation:

To solve this, we'll need to first differentiate the position function to get the velocity function, differentiate that function to get the acceleration function, and then plug our specific time into that funtion to get the acceleration at that moment. Differentiating the position equation term-by-term, using the power rule, which states that

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

and the constant rule, which states that the derivative of a constant is 0, we get

$\begin{align*}v(t)&= p'(t) \\ p(t)&=t^3+3t^2-15t+6 \\ p'(t) &= 3t^2+(2\cdot 3)t^{2-1}-15t^{1-1}+0 \\ p'(t)&=3t^2+6t-15 \\v(t) &=3t^2+6t-15\end{align*}$

Applying the same rules to this equation gives us

$\begin{align*}a(t)&= v'(t) \\ v(t)&=3t^2+6t-15 \\ v'(t) &= (2\cdot 3)t^{2-1}+6t^{1-1}+0 \\ v'(t)&=6t+6 \\a(t) &=6t+6\end{align*}$

Evaluating at t=3 gives us

$\begin{align*} a(1)&= 6(1)+6 \\ &=6+6\\&=12\,m/s \end{align*}$

Report an Error

View Calculus Tutors

Samarth
Certified Tutor

Maharaja Sayajirao University of Baroda, Bachelor of Engineering, Mechanical Engineering. Concordia University-Seward, Master...

View Calculus Tutors

Muhammad
Certified Tutor

The University of Texas at Dallas, Bachelor of Science, Mechanical Engineering.

View Calculus Tutors

John
Certified Tutor

The University of Texas at Arlington, Bachelor of Science, Mathematics. Stanford University, Master of Science, Aerospace Eng...

All Calculus 1 Resources

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

Popular Subjects

GMAT Tutors in Chicago, LSAT Tutors in Chicago, Spanish Tutors in Boston, Computer Science Tutors in Houston, French Tutors in Phoenix, Statistics Tutors in Phoenix, Algebra Tutors in Dallas Fort Worth, ISEE Tutors in Dallas Fort Worth, Biology Tutors in Houston, SAT Tutors in Washington DC

Popular Courses & Classes

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

Popular Test Prep

LSAT Test Prep in Chicago, SSAT Test Prep in New York City, GRE Test Prep in Boston, GMAT Test Prep in Seattle, MCAT Test Prep in Chicago, GRE Test Prep in Los Angeles, GMAT Test Prep in San Francisco-Bay Area, LSAT Test Prep in Miami, ISEE Test Prep in Houston, GRE Test Prep in San Diego

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 : Acceleration

Study concepts, example questions & explanations for Calculus 1

All Calculus 1 Resources

Example Questions

Example Question #291 : How To Find Acceleration

Example Question #291 : Acceleration

Example Question #292 : Acceleration

Example Question #683 : Spatial Calculus

Example Question #293 : Acceleration

Example Question #294 : Acceleration

Example Question #295 : Acceleration

Example Question #296 : Acceleration

Example Question #297 : Acceleration

Example Question #298 : Acceleration

All Calculus 1 Resources

Report an issue with this question

DMCA Complaint

Contact Information

Address

Complaint Details

Email address:
Your name:
Feedback: