Call Now to Set Up Tutoring:

(888) 888-0446

Calculus 1 : Distance

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 #771 : Spatial Calculus

The velocity of an object is given by the equation v(t) = 3t^2 - 5 . What is the distance covered by the object from time t = 2 to time t = 3 ?

Possible Answers:

None of these.

Correct answer:

Explanation:

The distance covered by the object can be found by integrating the velocity equation from time t= 2 to time t = 3 . To integrate the velocity equation we can use the power rule where if

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

Using this rule, the distance is calculated as

$x = \int_2^3 (3t^2 - 5)dt = \frac{3t^{(2+1)}}{2+1} - \frac{5t^{(0+1)}}{0+1}]_2^3 = t^3 - 5t]_2^3$

x = (3)^3 - 5(3) - (2)^3 + 5(2) = 27 - 15 - 8 +10 = 14

Report an Error

Example Question #771 : Calculus

The velocity of an object is given by the equation $v(t) = \frac{1}{2}\cos(t)$ . What is the distance travelled by the object from time t= 0 to time t = 1 ?

Possible Answers:

$x = \frac{\sin(1)}{2}$

x =1

$x = \frac{\sin(1)}{2} - \frac{1}{2}$

$x = \frac{\sin(1)}{2} - \frac{1}{4}$

Correct answer:

$x = \frac{\sin(1)}{2}$

Explanation:

The distance travelled by the object can be found by integrating the velocity equation.

Because the derivative of $\sin(t)$ is $\cos(t)$ , the integral of $\cos(t)$ is $\sin(t)$ .

Therefore

$x = \int_0^1 (\frac{1}{2}\cos(t))dt = \frac{1}{2}\sin(t)]_0^1 = \frac{1}{2}\sin(1) - \frac{1}{2}\sin(0) = \frac{\sin(1)}{2} - 0$

$x = \frac{\sin(1)}{2}$

Report an Error

Example Question #771 : Spatial Calculus

The velocity of an object is given by the equation v(t) = 2t - 6t^2 . What is the distance travelled by the object from t= 1 to t = 2 ?

Possible Answers:

Correct answer:

Explanation:

The distance travelled by the object can be found by integrating the position equation for the object from t =1 to t =2 . 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 distance travelled by the object is

$s = \int_1^2 (2t - 6t^2)dt = \frac{2t^{(1+1)}}{1+1} - \frac{6t^{(2+1)}}{2+1}]_1^2 = t^2 - 2t^3]_0^1$ .

s = (2)^2 - 2(2)^3 - (1)^2 + 2(1)^3= 4 - 16 - 1 + 2 = -11

Report an Error

Example Question #42 : How To Find Distance

The velocity of an object is v(t) = 4t -7 . What is the distance travelled by the object from t= 0 to t = 3 ?

Possible Answers:

Correct answer:

Explanation:

The distance travelled can be found by integrating the velocity from t= 0 to t =3 .

The velocity can be integrated using the power rule where

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

Applying this to the velocity equation gives

$\\s = \int_0^3 (4t -7t^0)dt = \frac{4t^{(1+1)}}{1+1} -\frac{7t^{(0+1)}}{0+1}]_0^3 = 2t^2 - 7t]_0^3 \Rightarrow \\ \\ \Rightarrow 2(3)^2 - 7(3) - 2(0)^2 + 7(0) = 18 - 21 - 0 + 0 = -3$

Report an Error

Example Question #41 : Distance

The position of an object is given by the equation x(t) = 2t^3 - 3t +4 . What is the distance from the object's location at t = 1 to the object's location at t =3 ?

Possible Answers:

Correct answer:

Explanation:

To find the distance between the two locations we can subtract the position at t = 1 from the position at t = 3 .

$\\x(3) - x(1) = 2(3)^3 - 3(3) +4 - 2(1)^3 + 3(1) - 4 \\= 54 - 9 +4 -2 +3 -4 \\= 46$

Therefore the distance between the locations is .

Report an Error

Example Question #43 : How To Find Distance

The velocity of an object is given by the equation v(t) = 300t - 20t^3 +500 . What is the distance covered by the object from t = 0 to t = 10 ?

Possible Answers:

6,450

-30,000

70,000

30,000

Correct answer:

-30,000

Explanation:

To find the distance travelled we can integrate the velocity equation 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 this equation we find that,

$s = \int_0^{10} (300t-20t^3+500t^0)dt = \frac{300t^{1+1}}{1+1} - \frac{20t^{3+1}}{3+1} +\frac{500t^{0+1}}{0+1}]_0^{10}$

s = 150(10)^2 - 5(10)^4 + 500(10) -150(0)^2 - 5(0)^4 +500(0)

s = 15,000 - 50,000 +5,000 = -30,000 .

Report an Error

Example Question #781 : Calculus

The velocity of an object is given by the equation v(t) = 8t^3 . What is the distance covered by the object from t = 0 to t = 2 ?

Possible Answers:

None of these.

Correct answer:

Explanation:

To find the distance covered by the object we can integrate the velocity equation. This can be done using the power rule where if

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

Therefore the distance covered by the object is

$s = \int_0^2 (8t^3)dt = \frac{8t^{(3+1)}}{3+1}]_0^2 = 2t^4]_0^2 = 2(2)^4 - 2(0)^4 = 32$ .

Report an Error

Example Question #45 : How To Find Distance

The acceleration of an object is given by the equation a(t) = 6t . What is the distance covered by the object from time t =0 to t =1 , if the initial velocity of the object is ?

Possible Answers:

Correct answer:

Explanation:

The distance covered by the object can be found by integrating the acceleration twice. 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 gives

$v(t) = \int (6t)dt = \frac{6t^{(1+1)}}{1+1} + C = 3t^2 +C$ .

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

v(0) = 3(0)^2 + C = 0 +C = 3

Therefore C = 3 and v(t) = 3t^2 + 3 .

Integrating the velocity equation from t= 0 to t= 1 will give us the distance covered by the object.

$s= \int_0^1 (3t^2+3)dt = \frac{3t^{(2+1)}}{2+1} + \frac{3t^{(0+1)}}{0+1}]_0^1 = t^3 + 3t]_0^1$

s = (1)^3 + 3(1) - (0)^3 - 3(0) = 1 + 3 -0 - 0= 4

Report an Error

Example Question #781 : Spatial Calculus

The velocity of an object is given by the equation v(t) = 30t^2 - 4 . What is the distance covered by the object between t= 0 and t =1 ?

Possible Answers:

Correct answer:

Explanation:

The distance covered by the object can be found by integrating the velocity from t = 0 to t =1 , using the power rule where if

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

Applying this to the velocity equation gives

$s = \int_0^1 (30t^2-4)dt = \frac{30t^{2+1}}{2+1} - \frac{4t^{0+1}}{0+1}]_0^1 = 10t^3 - 4t]_0^1$

s = 10(1)^3 - 4(1) - 10(0)^3 +4(0) = 10 -4 - 0 + 0 = 6

Report an Error

Example Question #45 : Distance

The velocity equation of an object is given by the equation v(t) = 20t - 3t^2 . What is the distance covered by the object from t= 1 to t =2 ?

Possible Answers:

Correct answer:

Explanation:

The distance covered can be found by integrating the velocity from t= 1 to t =2 using the power rule, where if

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

Therefore the distance covered is

$s = \int_1^2 (20t - 3t^2)dt = \frac{20t^{(1+1)}}{1+1} - \frac{3t^{2+1}}{2+1}]_1^2 = 10t^2 - t^3]_1^2$

s= 10(2)^2 - (2)^3 - 10(1)^2 + (1)^3 = 40 - 8 - 10 + 1 = 23

Report an Error

View Calculus Tutors

Jesus Ricardo
Certified Tutor

Benemerita Universidad Autónoma de Puebla, Bachelor, Physics.

View Calculus Tutors

Mayowa
Certified Tutor

Logan University, Bachelor of Science, Biology, General.

View Calculus Tutors

Luis
Certified Tutor

State University of New York at New Paltz, Doctor of Science, Health Physics.

All Calculus 1 Resources

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

Popular Subjects

Calculus Tutors in Seattle, GRE Tutors in Boston, Biology Tutors in Boston, SAT Tutors in Chicago, Math Tutors in Atlanta, Computer Science Tutors in Washington DC, GMAT Tutors in Phoenix, ACT Tutors in Philadelphia, English Tutors in New York City, SSAT Tutors in Seattle

Popular Courses & Classes

ACT Courses & Classes in Chicago, LSAT Courses & Classes in Seattle, GMAT Courses & Classes in Phoenix, MCAT Courses & Classes in Denver, ISEE Courses & Classes in Houston, MCAT Courses & Classes in Atlanta, ACT Courses & Classes in Dallas Fort Worth, GRE Courses & Classes in Houston, GMAT Courses & Classes in Boston, SSAT Courses & Classes in Atlanta

Popular Test Prep

SSAT Test Prep in Dallas Fort Worth, MCAT Test Prep in San Francisco-Bay Area, SSAT Test Prep in San Diego, GRE Test Prep in San Diego, SSAT Test Prep in Phoenix, SSAT Test Prep in Philadelphia, MCAT Test Prep in Los Angeles, MCAT Test Prep in Denver, SAT Test Prep in Chicago, LSAT Test Prep in New York City

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 : Distance

Study concepts, example questions & explanations for Calculus 1

All Calculus 1 Resources

Example Questions

Example Question #771 : Spatial Calculus

Example Question #771 : Calculus

Example Question #771 : Spatial Calculus

Example Question #42 : How To Find Distance

Example Question #41 : Distance

Example Question #43 : How To Find Distance

Example Question #781 : Calculus

Example Question #45 : How To Find Distance

Example Question #781 : Spatial Calculus

Example Question #45 : Distance

All Calculus 1 Resources

Report an issue with this question

DMCA Complaint

Contact Information

Address

Complaint Details

Email address:
Your name:
Feedback: