We are open Saturday and Sunday!

Call Now to Set Up Tutoring:

(888) 888-0446

Calculus 1 : How to find 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

← Previous 1 2 3 4 5 6 7 8 9 10 11 12 Next →

Example Question #111 : How To Find Distance

Find the distance function given the following velocity function:

v(x)=2x-1

Possible Answers:

d(x)=x^2-x+C

d(x)=0

d(x)=x^2-x

d(x)=2

Correct answer:

d(x)=x^2-x+C

Explanation:

To solve, you must integrate v(t) to find d(t) once using the power rule for integrals.

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

Thus,

$d(x)=\int v(x)dx=\int (2x-1)dx=x^2-x+C$

Report an Error

Example Question #112 : How To Find Distance

The velocity of a particle is given by the function $v(t)=\frac{4}{cos^2(2t)}$ . Find the distance traveled by the particle over the interval of time t=[1,2] .

Possible Answers:

11.6

6.7

13.4

5.8

2.9

Correct answer:

6.7

Explanation:

Velocity is the time derivative of position, and by that token position can be found by integrating a known velocity function with respect to time:

$p(t)=\int v(t)dt$

Now, if this integral were to be taken over an interval of time t=[a,b] , this will give a finite value, a change in position, i.e. a distance travelled:

$d=\int_a^b v(t)dt$

For the velocity function

$v(t)=\frac{4}{cos^2(2t)}$

The distance travelled can be found via knowledge of the following derivative properties:

Trigonometric derivative:

$d[tan(u)]=\frac{du}{cos^2(u)}$

The distance travelled over t=[1,2] is:

d=2tan(2t)|_1^2

d=(2tan(2(2)))-2tan(2(1))=6.7

Report an Error

Example Question #113 : How To Find Distance

The velocity of a particle is given by the function v(t)=3sin^2(t)cos(t) . Find the distance traveled by the particle over the interval of time $t=[0,\frac{\pi}{4}]$ .

Possible Answers:

$\frac{3\sqrt{2}}{4}$

$\frac{\sqrt{2}}{2}$

$\frac{\sqrt{2}}{8}$

$\frac{\sqrt{2}}{4}$

$\frac{3\sqrt{2}}{2}$

Correct answer:

$\frac{\sqrt{2}}{4}$

Explanation:

Velocity is the time derivative of position, and by that token position can be found by integrating a known velocity function with respect to time:

$p(t)=\int v(t)dt$

Now, if this integral were to be taken over an interval of time t=[a,b] , this will give a finite value, a change in position, i.e. a distance travelled:

$d=\int_a^b v(t)dt$

For the velocity function

v(t)=3sin^2(t)cos(t)

The distance travelled can be found via knowledge of the following derivative properties:

Trigonometric derivative:

d[sin(u)]=cos(u)du

w=sin(t);dw=cos(t)

$\int 3w^2dw=w^3=sin^3(t)$

The distance travelled over $t=[0,\frac{\pi}{4}]$ is:

$d=sin^3(t)|_0^{\frac{\pi}{4}}$

$d=sin^3(\frac{\pi}{4})-sin^3(0)=\frac{\sqrt{2}}{4}$

Report an Error

Example Question #114 : How To Find Distance

The velocity of a particle is given by the function v(t)=4t+8 . Find the distance traveled by the particle over the interval of time t=[1,3] .

Possible Answers:

Correct answer:

Explanation:

Velocity is the time derivative of position, and by that token position can be found by integrating a known velocity function with respect to time:

$p(t)=\int v(t)dt$

Now, if this integral were to be taken over an interval of time t=[a,b] , this will give a finite value, a change in position, i.e. a distance travelled:

$d=\int_a^b v(t)dt$

For the velocity function

v(t)=4t+8

The distance travelled over t=[1,3] is:

d=(2t^2+8t)|_1^3

d=(2(3)^2+8(3))-(2(1)^2+8(1))=32

Report an Error

Example Question #111 : Distance

The velocity of a particle is given by the function v(t)=8sin(t)cos(t) . Find the distance traveled by the particle over the interval of time $t=[0,\frac{\pi}{2}]$ .

Possible Answers:

Correct answer:

Explanation:

Velocity is the time derivative of position, and by that token position can be found by integrating a known velocity function with respect to time:

$p(t)=\int v(t)dt$

Now, if this integral were to be taken over an interval of time t=[a,b] , this will give a finite value, a change in position, i.e. a distance travelled:

$d=\int_a^b v(t)dt$

For the velocity function

v(t)=8sin(t)cos(t)

The distance travelled can be found via knowledge of the following derivative properties:

Trigonometric derivative:

d[sin(u)]=cos(u)du

w=sin(t);dw=cos(t)

$\int8wdw=4w^2$

The distance travelled over $t=[0,\frac{\pi}{2}]$ is:

$d=4sin^2(t)|_0^{\frac{\pi}{2}}$

$d=(4sin^2(\frac{\pi}{2}))-4sin^2(0)=4$

Report an Error

Example Question #111 : Distance

Find the distance from points: (-3,-4) to (5,-5)

Possible Answers:

$\sqrt{65}$

$\sqrt{106}$

Correct answer:

$\sqrt{65}$

Explanation:

This is simply the application of the distance formula:

The distance is going to be equal to:

$d=\sqrt{(y_1-y_0)^2+(x_1-x_0)^2}=\sqrt{(-4+5)^2+(5+3)^2}=\sqrt{65}$

Report an Error

Example Question #117 : How To Find Distance

Determine the distance travelled by a person if their displacement is 100m , and they moved equal lengths West and North, and didn't move in any other direction.

Possible Answers:

$100\sqrt{2}m$

100m

$50\sqrt{2}m$

$200\sqrt{2}m$

Correct answer:

$100\sqrt{2}m$

Explanation:

We know that displacement is just the magnitude of the vector moving from point to point . Distance however is the sum of the movements. In this question, we can treat the distance as the hypotenuse of a triangle, and the movements in the west and north directions as the 2 legs. Since the person moved the same distance in each direction, which I will call , we can determine it by doing:

100^2= a^2+a^2

$\sqrt{\frac{100^2}{2}}=50\sqrt{2}=a$

Since the person walked in both directions, the total distance travelled is:

$50\sqrt{2}+50\sqrt{2}=100\sqrt{2}=a$

Report an Error

Example Question #112 : Distance

Determined the displacement of a person who walks 30m west and 70m north.

Possible Answers:

$20\sqrt{10}$

$10\sqrt{58}m$

100m

-40m

Correct answer:

$10\sqrt{58}m$

Explanation:

The displacement in orthogonal directions(North and West) can be determined by using the pythagorean theorem:

$d=\sqrt{30^2+70^2}=\sqrt{5800}=10\sqrt{58}$

Report an Error

← Previous 1 2 3 4 5 6 7 8 9 10 11 12 Next →

View Calculus Tutors

Gregory
Certified Tutor

Pennsylvania State University-Main Campus, Bachelor of Science, Chemical Engineering. Carnegie Mellon University, Doctor of P...

View Calculus Tutors

Raphael
Certified Tutor

The University of Texas at Dallas, Bachelor of Science, Computer Science.

View Calculus Tutors

Chase
Certified Tutor

University of Memphis, Bachelor in Arts, Psychology.

All Calculus 1 Resources

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

Popular Subjects

Statistics Tutors in Atlanta, LSAT Tutors in Denver, Calculus Tutors in Boston, Calculus Tutors in Houston, Spanish Tutors in San Diego, GMAT Tutors in Seattle, MCAT Tutors in Atlanta, Algebra Tutors in Boston, SSAT Tutors in Atlanta, GMAT Tutors in Atlanta

Popular Courses & Classes

GMAT Courses & Classes in Seattle, ACT Courses & Classes in New York City, Spanish Courses & Classes in San Diego, MCAT Courses & Classes in Miami, ACT Courses & Classes in Atlanta, SSAT Courses & Classes in Phoenix, GRE Courses & Classes in New York City, SAT Courses & Classes in Phoenix, ISEE Courses & Classes in Phoenix, LSAT Courses & Classes in San Francisco-Bay Area

Popular Test Prep

LSAT Test Prep in Houston, ACT Test Prep in Miami, MCAT Test Prep in Atlanta, ACT Test Prep in Phoenix, ISEE Test Prep in Seattle, LSAT Test Prep in New York City, SSAT Test Prep in Houston, GMAT Test Prep in Philadelphia, GRE Test Prep in Seattle, GRE Test Prep in San Francisco-Bay Area

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 distance

Study concepts, example questions & explanations for Calculus 1

All Calculus 1 Resources

Example Questions

Example Question #111 : How To Find Distance

Example Question #112 : How To Find Distance

Example Question #113 : How To Find Distance

Example Question #114 : How To Find Distance

Example Question #111 : Distance

Example Question #111 : Distance

Example Question #117 : How To Find Distance

Example Question #112 : Distance

All Calculus 1 Resources

Report an issue with this question

DMCA Complaint

Contact Information

Address

Complaint Details

Email address:
Your name:
Feedback: