Call Now to Set Up Tutoring:

(888) 888-0446

Calculus 1 : Spatial 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 #51 : Velocity

The position of an object is given by the equation $x(t) = \cos^2(t)$ . What is the velocity of the object when t =0 ?

Possible Answers:

$\frac{\sqrt2}{3}$

$\frac{1}{2}$

Correct answer:

Explanation:

Given the equation $x(t) = \cos^2(t)$ , the velocity can be found by differentiating the position equation. To properly differentiate this equation, we must use the chain rule where if

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

Therefore, the derivative of the function is,

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

We now use t = 0 to find the velocity.

$v(0) = -2\cos(0)\sin(0) = -2(1)(0) = 0$

Report an Error

Example Question #52 : Spatial Calculus

The acceleration of an object is given by the equation a(t) =40 . What is the velocity at t = 3 if the object has an initial velocity of ?

Possible Answers:

140

120

100

Correct answer:

140

Explanation:

The velocity of the object is the integral of the accelertion. We can use the power rule to integrate the equation. The power rule is that

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

Therefore

$v(t) = \int (40)dt = \int (40t^0)dt = \frac{40t^{(0+1)}}{0+1} +C = 40t^1 +C$

To find the velocity we must now solve for the value of , using the initial velocity of the object.

v(0) = 40(0) +C = 0 + C= 20

Therefore C = 20 and v(t) = 40t + 20

We can now find the velocity at time t = 3

v(3) = 40(3) +20 = 120 +20 = 140

Report an Error

Example Question #51 : How To Find Velocity

The position of an object is given by the equation $x(t) = \sin^2(t) + \cos^2(t)$ . What is the velocity equation of the object?

Possible Answers:

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

v(t) = 0

$v(t) = -\sin(t)\cos(t)$

$v(t) = \sin(t)\cos(t)$

Correct answer:

v(t) = 0

Explanation:

The velocity of the object can be found by differentiating the position equation. To differentiate the position equation 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 where

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

Therefore the velocity equation is

$v(t) = x'(t) = 2\sin(t)\cos(t) + 2\cos(t)(-\sin(t)) \Rightarrow \\ \Rightarrow 2\sin(t)\cos(t) - 2\sin(t)\cos(t) = 0$

Report an Error

Example Question #2 : Calculus Review

The position of an object is given by the equation x(t) = e^t +t . What is the velocity of the object at t = 2 ?

Possible Answers:

15.5

8.3

9.3

Correct answer:

8.3

Explanation:

The velocity of the object can be found by differentiating the position equation of the object. To differentiate the position equation of the object, we can use the power rule for the second term where if

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

Using this rule we find that

$v(t) = x'(t) = e^t + (1)t^{(1-1)} = e^t +1$

We can now use the value of t = 2 to solve for the velocity at t = 2

$v(2) = e^2 +1 \approx 2.7^2 +1 = 8.3$

Report an Error

Example Question #53 : Velocity

The position of an object is given by the equation $x(t) = 7e^t - \cos^2(t)$ . What is the equation for the velocity of the object?

Possible Answers:

$v(t) = 7e^t - \sin(t)\cos(t)$

$v(t) = 7te^t + \sin(t)\cos(t)$

$v(t) = 7e^t + 2\sin(t)\cos(t)$

$v(t) = 7 - 2\sin(t)\cos(t)$

Correct answer:

$v(t) = 7e^t + 2\sin(t)\cos(t)$

Explanation:

The velocity of the object is the derivative of the position equation. To differentiate the position equation 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

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

Applying these two rules to the position equation gives us

$v(t) = x'(t) = 7e^t - 2\cos(t)(-\sin(t)) = 7e^t +2\sin(t)\cos(t)$

Report an Error

Example Question #51 : How To Find Velocity

The position of a car is defined as $p(t)=5t^{2}-3t+7$ . What is the car's velocity at t=3 ?

Possible Answers:

Correct answer:

Explanation:

The velocity v(t) is defined as the first derivative of its position p(t) , or v(t)=p'(t) .

Using the power rule to differentiate,

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

We find.

$p(t)=5t^{2}-3t+7$ , v(t)=p'(t)=10t-3 .

Swapping t=3 , v(t)=10(3)-3=30-3=27 .

Report an Error

Example Question #51 : Velocity

The position of a particle is defined as $p(t)=7t^{2}+4t+3$ . What is the particle's velocity at t=5 ?

Possible Answers:

Correct answer:

Explanation:

The velocity v(t) is defined as the first derivative of its position p(t) , or v(t)=p'(t) .

Using the power rule when differentiating,

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

we find,

$p(t)=7t^{2}+4t+3$ , v(t)=p'(t)=14t+4 .

Swapping t=5 , v(t)=14(5)+4=70+4=74 .

Report an Error

Example Question #58 : Spatial Calculus

The position of a particle is defined as $p(t)=3t^{2}+2t+10$ . What is the particle's velocity at t=1 ?

Possible Answers:

Correct answer:

Explanation:

The velocity v(t) is defined as the first derivative of its position p(t) , or v(t)=p'(t) .

Use the power rule to differentiate the position function.

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

Since $p(t)=3t^{2}+2t+10$ , v(t)=p'(t)=6t+2 .

Swapping t=1 , v(t)=6(1)+2=6+2=8 .

Report an Error

Example Question #54 : How To Find Velocity

The acceleration of an object is given by the equation a(t) = 12t^2 - 4t^3 . What is the velocity of the object at time t = 1 , if the velocity of the object is at time t = 3 ?

Possible Answers:

Correct answer:

Explanation:

The velocity of the object can be found by integrating the acceleration of the object. This can be done using the power rule where if

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

Integrating the acceleration equation gives us

$v(t) = \int (12t^2 - 4t^3)dt = \frac{12t^{(2+1)}}{2+1} - \frac{4t^{(3+1)}}{3+1} + C = 4t^3 - t^4 +C$

We can solve for the value of using the velocity of the object at t = 3

v(3) = 4(3)^3 - (3)^4 + C =108 - 81+C = 27+C = 3

Therefore C= -24 and v(t) = 4t^3 - t^4 -24

We can now find the velocity at time t =1

v(1) = 4(1)^3 - (1)^4 -24 = 4 - 1 -24 = -21

Report an Error

Example Question #52 : How To Find Velocity

The position of an object is given by the equation $x(t) = 13e^{3t} - 3$ . What is the velocity of the object?

Possible Answers:

$v(t) = 3e^{3t} - 3t$

$v(t) = 13e^{3t} - 3t$

$v(t) = 39e^{3t}$

$v(t) = 13e^{3t}$

Correct answer:

$v(t) = 39e^{3t}$

Explanation:

The velocity of the object can be found by differentiating the position equation of the object. To accurately 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 where if

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

Therefore the velocity of the object is

$v(t) = 13e^{3t}(3(1)t^{1-1}) - 3(0)t^{(0-1)} = 39e^{3t} - 0 = 39e^{3t}$

Report an Error

View Calculus Tutors

Samantha
Certified Tutor

Middlesex County College, Associate in Science, Sustainability Studies.

View Calculus Tutors

Satweka
Certified Tutor

The University of Texas at Dallas, Master of Science, Biomedical Engineering.

View Calculus Tutors

Aniket
Certified Tutor

Gujarat Technological University, Sarvajanik College of Engineering and Technology, Bachelor of Engineering, Computer Enginee...

All Calculus 1 Resources

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

Popular Subjects

Statistics Tutors in Philadelphia, Algebra Tutors in Phoenix, Calculus Tutors in Boston, ISEE Tutors in Houston, GMAT Tutors in Phoenix, SAT Tutors in Houston, LSAT Tutors in Los Angeles, Calculus Tutors in Phoenix, Computer Science Tutors in New York City, MCAT Tutors in Washington DC

Popular Courses & Classes

Spanish Courses & Classes in Seattle, GRE Courses & Classes in Philadelphia, Spanish Courses & Classes in San Diego, SAT Courses & Classes in San Francisco-Bay Area, LSAT Courses & Classes in New York City, SAT Courses & Classes in Denver, MCAT Courses & Classes in Philadelphia, MCAT Courses & Classes in Boston, SAT Courses & Classes in Houston, SAT Courses & Classes in Miami

Popular Test Prep

SSAT Test Prep in New York City, SAT Test Prep in Denver, ISEE Test Prep in Dallas Fort Worth, ISEE Test Prep in Los Angeles, SAT Test Prep in Houston, LSAT Test Prep in Phoenix, ACT Test Prep in New York City, MCAT Test Prep in Washington DC, SSAT Test Prep in Denver, ISEE Test Prep in Chicago

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 : Spatial Calculus

Study concepts, example questions & explanations for Calculus 1

All Calculus 1 Resources

Example Questions

Example Question #51 : Velocity

Example Question #52 : Spatial Calculus

Example Question #51 : How To Find Velocity

Example Question #2 : Calculus Review

Example Question #53 : Velocity

Example Question #51 : How To Find Velocity

Example Question #51 : Velocity

Example Question #58 : Spatial Calculus

Example Question #54 : How To Find Velocity

Example Question #52 : How To Find Velocity

All Calculus 1 Resources

Report an issue with this question

DMCA Complaint

Contact Information

Address

Complaint Details

Email address:
Your name:
Feedback: