We are open Saturday and Sunday!

Call Now to Set Up Tutoring:

(888) 888-0446

Calculus 1 : Position

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 #3 : Integration

The velocity equation of an object is given by the equation v(t) = (4t^2)(3t) . What is the position of the object at time t = 4 if the initial position of the object is ?

Possible Answers:

768

192

Correct answer:

768

Explanation:

The position of the object can be found by integrating the velocity equation and solving for x(4) . To integrate the velocity equation we first rewrite the equation.

v(t) = (4t^2)(3t) = 12t^3

To integrate this equation we must use the power rule where,

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

Applying this to the velocity equation gives us,

$x(t) = \int (12t^3)dt = \frac{12t^{3+1}}{3+1} = \frac{12t^4}{4} = 3t^4 +C$ .

We must solve for the value of by using the initial position of the object.

x(0) = 3(0)^4 + C =0 + C= 0

Therefore, C = 0 and x(t) = 3t^4 .

x(4) = 3(4)^4 = 3(256) = 768

Report an Error

Example Question #22 : How To Find Position

The velocity of an object is given by the equation $v(t)= 2\sin(t)$ . What is the equation for the position of the object if the object has an initial position of ?

Possible Answers:

$x(t) = -2\cos(t)$

$x(t) = 2\cos(t)$

$x(t) = -2\cos(t) +2$

$x(t) = -2\sin(t)$

Correct answer:

$x(t) = -2\cos(t) +2$

Explanation:

The position of the object can be found by integrating the equation for the object's velocity. Knowing that the derivative of $-\cos(t)$ is $\sin(t)$ , the integral of $\sin(t)$ must be $-\cos(t)$ .

Integrating the velocity equation gives us,

$x(t) = \int (2\sin(t))dt = 2(-\cos(t)) +C = -2\cos(t) +C$

To find the complete solution of the position equation we must use the initial position.

$x(0) = -2\cos(0) +C = -2 +C = 0$

Therefore C = 2 and $x(t) = -2\cos(t) +2$

Report an Error

Example Question #871 : Spatial Calculus

Find a vector perpendicular to (5,7) .

Possible Answers:

(7,-5)

(-5,7)

(5,-7)

(-7,5)

(-5,-7)

Correct answer:

(-7,5)

Explanation:

By definition, a vector (a,b) has a perpendicular vector (-b,a) .

Therefore, the vector (5,7) has a perpendicular vector (-7,5) .

Report an Error

Example Question #872 : Spatial Calculus

Find a vector perpendicular to (2,6) .

Possible Answers:

(2,-6)

(-2,6)

(-6,2)

(-2,-6)

(6,2)

Correct answer:

(-6,2)

Explanation:

By definition, a vector (a,b) has a perpendicular vector (-b,a) .

Therefore, the vector (2,6) has a perpendicular vector (-6,2) .

Report an Error

Example Question #25 : How To Find Position

Find a vector perpendicular to (3,12) .

Possible Answers:

(12,3)

(3,-12)

(-12,3)

(-3,-12)

(-3,12)

Correct answer:

(-12,3)

Explanation:

By definition, a vector (a,b) has a perpendicular vector (-b,a) .

Therefore, the vector (3,12) has a perpendicular vector (-12,3) .

Report an Error

Example Question #1 : Integration

The acceleration of an object is given by the equation a(t) = 4t+1 . What is the equation for the position of the object, if the object has an initial velocity of and an initial position of ?

Possible Answers:

$x(t) = \frac{2t^3}{3} + \frac{t^2}{2} + 5t$

x(t) = 2t^3 + t^2 + 10t

x(t) = 2t^3 - t^2 + 10t

$x(t) = \frac{2t^3}{3} + \frac{t^2}{2} + 10t$

Correct answer:

$x(t) = \frac{2t^3}{3} + \frac{t^2}{2} + 10t$

Explanation:

To find the position of the object we must use the power rule to integrate the acceleration equation twice. The power rule is such that

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

Therefore integrating the acceleration equation gives us

$v(t) = \int (4t+1)dt = \frac{4t^{1+1}}{1+1} +\frac{1t^{0+1}}{0+1} +C = 2t^2 + t +C$

We can solve for the value of by using the initial velocity of the object.

v(0) = 2(0)^2 +(0)+C = 10

Therefore C = 10 and v(t) = 2t^2 + t + 10

To find the position of the object we integrate the velocity equation.

$x(t) = \int (2t^2+t+10)dt = \frac{2t^{2+1}}{2+1} + \frac{t^{1+1}}{1+1} + \frac{10t^{0+1}}{0+1} +C = \frac{2t^3}{3}+\frac{t^2}{2}+10t +C$

We can solve for this new value of by using the object's initial position

$x(0) = \frac{2(0)^3}{3} + \frac{(0)^2}{2} + 10(0) +C = 0 + C = 0$

Therefore C = 0 and $x(t) = \frac{2t^3}{3} + \frac{t^2}{2} + 10t$

Report an Error

Example Question #5 : Integration

The velocity of an object is given by the equation v(t) = 6t -3 . What is the position of the object at time t= 4 if the object has a position of and time t = 1 ?

Possible Answers:

Correct answer:

Explanation:

To find the position of the object we must first find the position equation of the object. The position equation can be found by integrating the velocity equation. This can be done using the power rule where if

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

Using this rule we find that

$x(t) = \int (6t-3)dt = \frac{6t^{1+1}}{1+1} - \frac{3t^{0+1}}{0+1} +C = 3t^2 - 3t +C$

Using the position of the object at time t = 1 we can solve for

x(1) = 3(1)^2 - 3(1) + C = 3 - 3 + C =0+C= 6

Therefore C = 6 and x(t) = 3t^2 - 3t +6

We can now find the position at time t = 4 .

x(4) = 3(4)^2 - 3(4) +6 = 48-12 +6 = 42

Report an Error

Example Question #21 : How To Find Position

The velocity of an object is given by the equation v(t) = 5t^4 + 3t^2 +5 . what is the position of the object at t= 3 , if the initial position of the object is ?

Possible Answers:

285

345

193

510

Correct answer:

285

Explanation:

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

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

Using the power rule the position of the object is

$x(t) = \int (5t^4 + 3t^2 + 5)dt = \frac{5t^{(4+1)}}{4+1} +\frac{3t^{(2+1)}}{2+1} + \frac{5t^{(0+1)}}{0+1} +C= t^5 +t^3 +5t +C$ .

The value of can be found using the initial position of the object.

x(0) = (0)^5 +(0)^3 + 5(0) +C = 0

Therefore C = 0 and x(t) = t^5 + t^3 +5t .

The position of the object at t = 3 can now be found,

x(3) = (3)^5 + (3)^3 + 5(3) =243+ 27+ 15 = 285 .

Report an Error

Example Question #882 : Spatial Calculus

The velocity of an object is given by the equation v(t) = 12t - 39 . What is the position of the object at time t = 2 , if the initial position of the object is ?

Possible Answers:

None of these.

Correct answer:

Explanation:

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

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

Therefore the position equation of the object is

$x(t) = \int (12t-39)dt = \frac{12t^{(1+1)}}{1+1} -\frac{39t^{(0+1)}}{0+1}+C = 6t^2 - 39t +C$

We must now solve for the constant . We can do this using the initial position of the object.

x(0) = 6(0)^2 - 39(0) +C = 0 - 0 + C = 0

Therefore C = 0 and x(t) = 6t^2 - 39t

We can now find the position of the object at t = 2 .

x(2) = 6(2)^2 - 39(2) = 24 - 78 = -54

Report an Error

Example Question #883 : Spatial Calculus

The velocity of an object is v(t) = 1 - 9t^2 . What is the position of the object when t = 1 , if the position of the object is at t =2 ?

Possible Answers:

Correct answer:

Explanation:

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

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

Therefore the position of the object is

$x(t) = \int (1t^0-9t^2)dt = \frac{1t^{(0+1)}}{0+1} + \frac{9t^{(2+1)}}{2+1}+C = t + 3t^3 + C$ .

We can find the value of using the position at t =2 .

x(2) = (2) - 3(2)^3 + C = 2-24+C=-22+C= 10

Therefore C= 32 and x(t) = t - 3t^3 + 32 .

x(1) = (1) - 3(1)^3 +32 = -2 + 32 = 30

Report an Error

View Calculus Tutors

Michelle
Certified Tutor

University of Nevada-Las Vegas, Bachelor of Science, Mathematics.

View Calculus Tutors

Leonard
Certified Tutor

Florida State University, Bachelor of Science, Environmental Science.

View Calculus Tutors

Mohammad
Certified Tutor

Rutgers University-New Brunswick, Bachelor of Science, Mathematics. Rutgers University-New Brunswick, Master of Science, Stat...

All Calculus 1 Resources

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

Popular Subjects

Algebra Tutors in Washington DC, Algebra Tutors in Atlanta, Calculus Tutors in Los Angeles, Physics Tutors in Atlanta, MCAT Tutors in Washington DC, GRE Tutors in Phoenix, GMAT Tutors in San Diego, GRE Tutors in San Diego, Reading Tutors in Denver, Spanish Tutors in Los Angeles

Popular Courses & Classes

LSAT Courses & Classes in Los Angeles, ACT Courses & Classes in Dallas Fort Worth, LSAT Courses & Classes in Atlanta, ISEE Courses & Classes in Los Angeles, SSAT Courses & Classes in Atlanta, SSAT Courses & Classes in Dallas Fort Worth, Spanish Courses & Classes in San Francisco-Bay Area, MCAT Courses & Classes in San Diego, ACT Courses & Classes in Seattle, SSAT Courses & Classes in Houston

Popular Test Prep

LSAT Test Prep in Atlanta, ACT Test Prep in Dallas Fort Worth, SAT Test Prep in Dallas Fort Worth, ISEE Test Prep in Atlanta, SAT Test Prep in Houston, MCAT Test Prep in Seattle, ISEE Test Prep in Houston, MCAT Test Prep in Washington DC, SSAT Test Prep in Houston, GRE Test Prep in Phoenix

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 : Position

Study concepts, example questions & explanations for Calculus 1

All Calculus 1 Resources

Example Questions

Example Question #3 : Integration

Example Question #22 : How To Find Position

Example Question #871 : Spatial Calculus

Example Question #872 : Spatial Calculus

Example Question #25 : How To Find Position

Example Question #1 : Integration

Example Question #5 : Integration

Example Question #21 : How To Find Position

Example Question #882 : Spatial Calculus

Example Question #883 : Spatial Calculus

All Calculus 1 Resources

Report an issue with this question

DMCA Complaint

Contact Information

Address

Complaint Details

Email address:
Your name:
Feedback: