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 #21 : Distance

Let B=(2a-x,y) . Let A=(x,y) . What is the distance between these two points?

Possible Answers:

2|x+y|

|x-y|

$\sqrt{2}|x-a|$

x+y

2|a|

Correct answer:

$\sqrt{2}|x-a|$

Explanation:

Let A=(x,y) and (2a-x,y) .

We will use the distance formula to find the distance d(A,B).

By definition we have:

$D(A,B)=\sqrt{(2x-2a)^2+(y-y)^2}=|x-a|\sqrt{2}$

Report an Error

Example Question #22 : Distance

What is the average function value of y=x^2 from the interval [1,3] ?

Possible Answers:

$\frac{13}{3}$

$\frac{8}{3}$

Correct answer:

$\frac{13}{3}$

Explanation:

Write the formula for average function value.

$f_{average}= \frac{1}{b-a}\int_{a}^{b} f(x)dx$

Substitute the function and the bounds.

$f_{average}= \frac{1}{3-1}\int_{1}^{3}x^2dx$

$f_{average}= \left(\frac{1}{2}\right)\left[\frac{1}{3}x^3\right]_{1}^{3}=\left(\frac{1}{2}\right)\left[\frac{1}{3}(3)^3-\frac{1}{3}(1)^3\right]$

$f_{average}= \frac{1}{2}\left[9-\frac{1}{3}\right]=\frac{1}{2}\left[\frac{26}{3}\right]=\frac{13}{3}$

Report an Error

Example Question #23 : Distance

An object is fired in the air, and the function which describes the object's height is y=-9x^2+8x+1 , where is in seconds. What is the height of the firing point to the maximum height of the object?

Possible Answers:

$\frac{25}{9}$

$\frac{16}{9}$

$\frac{5}{4}$

Correct answer:

$\frac{16}{9}$

Explanation:

Use the vertex formula to determine the location maximum height of the parabola.

$x=-\frac{b}{2a}=-\frac{8}{2(-9)}=\frac{4}{9}$

Substitute this value to the position function to find the height.

$y=-9\left(\frac{4}{9}\right)^2+8\left(\frac{4}{9}\right)+1$

$y=-9\left(\frac{16}{81}\right)+\frac{32}{9}+1$

$y=-\frac{16}{9}+\frac{32}{9}+\frac{9}{9}=\frac{25}{9}$

This is the maximum height, but at the start point x=0 , the y-intercept shows that the projectile is fired at a height of 1. Subtract the maximum from the height of the projectile.

$\frac{25}{9}-1=\frac{25}{9}-\frac{9}{9}= \frac{16}{9}$

The height from firing point to the maximum is $\frac{16}{9}$ .

Report an Error

Example Question #24 : Distance

Find the midpoint of the line segment connecting the points (0,3,-4) and (2,-6,3) .

Possible Answers:

(0.5,-2,-1)

(2,-3,-1)

(1,-1.5,-0.5)

(1,2,1)

Correct answer:

(1,-1.5,-0.5)

Explanation:

The coordinates of the midpoint is the average of the coordinates of each point.

$\left(\frac{x_1+x_1}{2},\frac{y_1+y_2}{2},\frac{z_1+z_2}{2}\right)$

Substituting in the values given:

$\left(\frac{0+2}{2},\frac{3+(-6)}{2},\frac{-4+3}{2}\right)$

$\left(\frac{2}{2},\frac{-3}{2},\frac{-1}{2}\right)$

Therefore the coordinates of the midpoint are (1,-1.5,-0.5) .

Report an Error

Example Question #25 : Distance

If the acceleration of an object is $f(t) = 2+\sin(t)$ . What is the displacement of the object from t=0 to t=2 , if the object had an initial velocity of ?

Possible Answers:

$4+\sin(2)$

$\sin(2)$

$4 - \sin(2)$

$8 - \sin(2)$

Correct answer:

$8 - \sin(2)$

Explanation:

The equation for displacement can be found by integrating the acceleration equation twice. Given the acceleration equation of $f(t)=2+\sin(t)$ .

The velocity equation is:

$v(t) = 2t - \cos(t) + C$

We can find the value of C using the initial velocity

v(0) = 0 -1 +C = 1

C = 2

The equation for velocity is then the integral of the acceleration function.

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

The equation of position is then

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

$x(t) = t^2 -\sin(t) +2t + C$

Solving for the distance between t=2 and t=0, we solve for x(2) and x(0).

$x(2) = 4 - \sin(2)+4+C$

x(0) = 0 - 0 + 0 + C

We can now subtract x(0) from x(2) to find our distance

$x(2) - x(0) = 4 - \sin(2)+4+ C - 0 +0 - 0 - C$

$x(2) - x(0) = 4 - \sin(2)+ 4 = 8-\sin(2)$

Report an Error

Example Question #761 : Spatial Calculus

Find the midpoint of the line segment between the two points (5,7,0) and (-1,3,6) .

Possible Answers:

(2,5,3)

(5,4,6)

(-4,10,6)

(4,10,6)

Correct answer:

(2,5,3)

Explanation:

The midpoint can be found by taking the average of each of the coordinates.

$\left(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2},\frac{z_1+z_2}{2}\right)$

Substituting in our values we find the midpoint as follows.

$\left(\frac{5-1}{2},\frac{7+3}{2},\frac{0+6}{2}\right) = (2,5,3)$

Report an Error

Example Question #27 : Distance

Find the distance from (2,-4,6) to point (1,-3,9) .

Possible Answers:

$\sqrt{59}$

$\sqrt{11}$

$\sqrt{15}$

$\sqrt{58}$

Correct answer:

$\sqrt{11}$

Explanation:

Write the distance formula.

$D=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2+(z_2-z_1)^2}$

Substitute the point values and solve for distance.

$D=\sqrt{(1-2)^2+(-3-(-4))^2+(9-6)^2}=\sqrt{1+1+9}$

$D=\sqrt{11}$

Report an Error

Example Question #762 : Spatial Calculus

What is the midpoint of the line segment between the two points (8,3,2) and (10,-3,-4) ?

Possible Answers:

(12,0,5)

(-2,6,-6)

(18,0,-2)

(9,0,-1)

Correct answer:

(9,0,-1)

Explanation:

The midpoint is the average of the coordinates.

Therefore, to find the midpoint, we must add each coordinate of the first point to each coordinate of the second point and divide by two, finding the halfway point between the two points.

$\left(\frac{8+10}{2},\frac{3+(-3)}{2},\frac{2+(-4)}{2}\right) = (9,0,-1)$

Report an Error

Example Question #21 : How To Find Distance

The velocity of a particle is given as v(t) = 8t -4 . What is the distance travelled by the particle t = 0 to t = 2 ?

Possible Answers:

Correct answer:

Explanation:

Given the velocity equation v(t) = 8t -4 , the position equation is the integral of the velocity from 0 to 2. To find this integral we can use the power rule.

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

Therefore, the integral of the velocity equation is

$x(t) = \int_{0}^{2}(8t-4)dt = 4t^2-4t]_{0}^{2}$ .

To evaluate this integral from t = 0 to t = 2 , we now substitute in the value for when t = 2 and substract the values for when t = 0 .

4(2)^2 - 4(2) - 4(0)^2 + 4(0) = 16 - 8 - 0 + 0 = 8

Report an Error

Example Question #30 : Distance

The velocity of an object is given by the equation v(t) = 9t^2 - 4t^3 . What is the distance travelled by the object from t = 2 to t = 3 ?

Possible Answers:

Correct answer:

Explanation:

Given the velocity equation v(t) = 9t^2 - 4t^3 , we can solve for the position equation by taking the integral of the velocity.

To do this we must use the power rule where if

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

Therefore, the integral of the velocity equation is

$x(t) = \int_{2}^{3}(9t^2-4t^3) = 3t^3 - t^4]_{2}^3$ .

We can now solve this by subtracting the value at t = 2 from the value when t = 3 .

$3t^3 - t^4]_{2}^3= 3(3)^3-(3)^4-3(2)^3+2^4 = 81 - 81 -24 +16 = -8$

Report an Error

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

View Calculus Tutors

Alexandra
Certified Tutor

University of Kent, Bachelor of Science, Mathematics.

View Calculus Tutors

Oselen
Certified Tutor

UNIVERSITY OF BENIN, Bachelor of Science, Mechanical Engineering. University of Windsor, Master of Science, Mechanical Engine...

View Calculus Tutors

Dilara
Certified Tutor

Hacettepe University, Bachelors, Mathematics. Balikesir University, Masters, Mathematics Education.

All Calculus 1 Resources

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

Popular Subjects

Chemistry Tutors in Boston, Reading Tutors in San Francisco-Bay Area, ACT Tutors in Miami, Reading Tutors in Dallas Fort Worth, SSAT Tutors in Miami, GRE Tutors in Denver, English Tutors in Seattle, ISEE Tutors in Denver, SAT Tutors in Los Angeles, SSAT Tutors in Philadelphia

Popular Courses & Classes

SAT Courses & Classes in Seattle, LSAT Courses & Classes in San Diego, MCAT Courses & Classes in Boston, SAT Courses & Classes in Houston, Spanish Courses & Classes in Boston, SAT Courses & Classes in Denver, MCAT Courses & Classes in Dallas Fort Worth, ISEE Courses & Classes in Seattle, SAT Courses & Classes in Boston, ACT Courses & Classes in Miami

Popular Test Prep

MCAT Test Prep in Washington DC, MCAT Test Prep in Miami, ISEE Test Prep in Dallas Fort Worth, LSAT Test Prep in Seattle, SAT Test Prep in Denver, SSAT Test Prep in Boston, LSAT Test Prep in Boston, SAT Test Prep in Seattle, ACT Test Prep in Miami, GMAT 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 : How to find distance

Study concepts, example questions & explanations for Calculus 1

All Calculus 1 Resources

Example Questions

Example Question #21 : Distance

Example Question #22 : Distance

Example Question #23 : Distance

Example Question #24 : Distance

Example Question #25 : Distance

Example Question #761 : Spatial Calculus

Example Question #27 : Distance

Example Question #762 : Spatial Calculus

Example Question #21 : How To Find Distance

Example Question #30 : Distance

All Calculus 1 Resources

Report an issue with this question

DMCA Complaint

Contact Information

Address

Complaint Details

Email address:
Your name:
Feedback: