Call Now to Set Up Tutoring:

(888) 888-0446

Calculus 1 : Velocity

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 #81 : Spatial Calculus

The position of an object is given by the function $x(t) = \sin(6t^3)$ . What is the velocity of the object at t= 0 ?

Possible Answers:

Correct answer:

Explanation:

To find the velocity of the object we can differentiate the position of the object. This can be done using the power rule where if

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

Using this rule we find that

$v(t) = \cos(6t^3)(6(3)t^{(3-1)}) = 18t^{2}\cos(6t^3)$ .

Therefore at t =0 the velocity of the object is

$v(0) = 18(0)^2\cos(6(0)^3) = (0)\cos(0) = 0$ .

Report an Error

Example Question #82 : Spatial Calculus

The position of an object is given by the equation $\sin(t)\cos^2(t)$ . What is the velocity of the object at t = 1 ?

Possible Answers:

$\cos^3(1) - 2\cos(1)\sin^2(1)$

$\cos^3(1) + 2\sin^2(1)\cos(1)$

$-\cos^3(1) + 2\sin(1)\cos^2(1)$

$\cos^3(1) + 2\sin(1)\cos^2(1)$

Correct answer:

$\cos^3(1) - 2\cos(1)\sin^2(1)$

Explanation:

The velocity of the object can be found by differentiating the position. This can be done using the power rule, the chain rule, and the product rule.

The power rule is

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

The chain rule is

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

The product rule is

$f(x)= j(x)k(x) \rightarrow f'(x) = j'(x)k(x) + j(x)k'(x)$ .

Appying these rules to the position equation gives us

$v(t) = (\cos(t))\cos^2(t) + \sin(t)(2\cos(t)(-\sin(t))) = \cos^3(t) - 2\cos(t)\sin^2(t)$ .

Using the velocity equation we find the velocity at t =1 to be

$v(1) = \cos^3(1) - 2\cos(1)\sin^2(1)$

Report an Error

Example Question #83 : Spatial Calculus

The position of an object is given by the equation $x(t) = -\cos(3t)$ . What is the velocity of the function at time t = 1 ?

Possible Answers:

$3\sin(1)$

$\sin(3)$

$3\sin(3)$

Correct answer:

$3\sin(3)$

Explanation:

The velocity if the function can be found by differentiating the position equation. This can be done using the power rule and the chain rule where if

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

and if

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

Using these rules we obtain

$v(t) = -(-\sin(3t))(3(1)t^{(1-1)}) = 3\sin(3t)$ .

We can now find the velocity at t = 1 .

$v(1) = 3\sin(3(1)) = 3\sin(3)$

Report an Error

Example Question #84 : Spatial Calculus

The acceleration of the an object is given by the equation a(t) = 10t^4 + 6t^5 - 6t . What is the velocity of the object if the object has an initial velocity of ?

Possible Answers:

v(t) = 2t^5 + t^6 + 3t^2

v(t) = 2t^5 + t^5 - 3t^2

v(t) = 2t^5 + 3t^6 + 3t^2

v(t) = 2t^5 + t^6 - 3t^2

Correct answer:

v(t) = 2t^5 + t^6 - 3t^2

Explanation:

The velocity of the object can be found by integrating the acceleration 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 get the equation

$v(t) = \int (10t^4 + 6t^5 - 6t)dt = \frac{10t^{(4+1)}}{4+1} + \frac{6t^{(5+1)}}{5+1} - \frac{6t^{(1+1)}}{1+1} +C$

v(t) = 2t^5 + t^6 - 3t^2 +C .

Using the initial velocity of the object we can find the value of .

v(0) = 2(0)^5 + (0)^6 - 3(0)^2 +C = 0 + 0 -0 +C = 0

Therefore C = 0 and v(t) = 2t^5 + t^6 - 3t^2 .

Report an Error

Example Question #85 : Calculus

Below is the graph of an object's position over time. Place the following in order from largest to smallest:

The instantaneous velocity at t=0

II. The average velocity between t=0.8 and t=2.2

III. The average velocity between t=3.12 and t=4.04

IV. The instantaneous velocity at t=2.2

Graph4

Possible Answers:

I, IV, III, II

I, II, IV, III

I, III, IV, II

II, IV, I, III

I, IV, II, III

Correct answer:

I, II, IV, III

Explanation:

The average velocity of an object over a given interval is given by $\frac{\bigtriangleup position }{\bigtriangleup time}$ and is represented on a position graphs by the slope of the secant line. The instantaneous velocity at a given time is represented by the slope of the tangent line at a point. Notice that the velocity is positive and decreasing. It is clear that is largest and is greater than III . Additionally, since the velocity is decreasing, is greater than . Hence we have: I, II, IV, III .

Report an Error

Example Question #85 : Spatial Calculus

A football is launched into the air during kick off. It's position is given by the function below. Find its instantaneous velocity at the time when it is caught by the kick returner. Assume the kick returner catches the ball at chest-level (about 4 feet off of the ground.)

p(t)=-16t^2+64t+2

Possible Answers:

v=-127ft/s

v=63ft/s

v=-63ft/s

v=0ft/s

Correct answer:

v=-63ft/s

Explanation:

The question asks us to find the slope of the tangent line at the time when the ball is caught. We can either use the limit definition of the derivative to find the slope directly or take the derivative, then plug in the appropriate time value to get the slope. I will be using the second method as it is a bit simpler.

We begin by solving for the time at which the ball is caught.

$-16t^2+64t+2=4 \implies t=.031s,3.97s$

Unless new rules come out for onsides kicks, we can choose the larger time value. Then we take the derivative:

p'(t)=-32t+64

We substitute the appropriate time value in to get:

p'(3.97)=-63ft/s

Report an Error

Example Question #81 : Spatial Calculus

Given the velocity function v=3t^2+2 , what is the average velocity from t=1 to t=2 ?

Possible Answers:

Correct answer:

Explanation:

Write the formula for average velocity.

$V_{avg}=\frac{X_f-X_i}{t_f-t_i}$

Integrate the velocity function to determine the position function.

$x(t)=\int v dt=\int (3t^2+2) \:dt$

x(t)= t^3+2t

Substitute the value of t=1 to determine X_i , or initial position.

x(1)= 1^3+2(1) =1+2=3

Substitute the value of t=2 to determine X_f , or final position.

x(2)= 2^3+2(2) =8+4=12

Substitute all the known values into the average velocity formula.

$V_{avg}=\frac{X_f-X_i}{t_f-t_i}= \frac{12-3}{2-1}=9$

Report an Error

Example Question #81 : Velocity

The position of an object is given by the function $x(t) = \sin(t)\cos(t)$ . What is the velocity of the object at t =3 ?

Possible Answers:

$\cos^2(3) + \sin^2(3)$

$\frac{\cos^2(3) - \sin^2(3)}{\cos^2(3)}$

$\cos^2(3) - \sin^2(3)$

None of these.

Correct answer:

$\cos^2(3) - \sin^2(3)$

Explanation:

The velocity of the object can be found by differentiating the position using the product rule where if

$f(x)= j(x)k(x) \rightarrow f'(x) = j'(x)k(x) + j(x)k'(x)$ .

Therefore the velocity of the object is

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

We can now solve for the velocity of the object at t =3 .

$v(3) = \cos^2(3) - \sin^2(3)$

Report an Error

Example Question #86 : Spatial Calculus

The position of an object is given by the equation x(t) = 10 . What is the velocity of the object at t =3 ?

Possible Answers:

Correct answer:

Explanation:

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

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

Then the velocity of the object is

$v(t) = 10(0)t^{(0-1)} = 0$ .

Therefore the velocity of the object is a constant .

Report an Error

Example Question #87 : Spatial Calculus

The acceleration of an object is given by the equation a(t) =8t^3 + 12t . What is the velocity of the equation at t =1 , if the velocity is at t =3 ?

Possible Answers:

Correct answer:

Explanation:

The velocity of the object can be found by differentiating the acceleration of the object 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

$v(t) = \int (8t^3 + 12t)dt = \frac{8t^{(3+1)}}{3+1} + \frac{12t^{1+1}}{1+1} + C = 2t^4 + 6t^2 +C$ .

We can find the value of using the velocity at t =3

v(3) = 2(3)^4 + 6(3)^2 +C = 162 + 54 +C = 216 +C = 0 .

Therefore C = -216 and v(t) = 2(t)^4 + 6t^2 - 216 .

We can now solve for the velocity of the object at t =1 .

v(1) = 2(1)^4 +6(1)^2 - 216 = 2 + 6 - 216 = -208

Report an Error

View Calculus Tutors

Dossevi
Certified Tutor

cole Ste Genevive Versailles, Bachelor of Science, Mathematics. Institut National Polytechnique de Grenoble, Master of Scienc...

View Calculus Tutors

Travis
Certified Tutor

Delaware State University, Bachelor of Science, Mathematics.

View Calculus Tutors

Shiblu
Certified Tutor

University of Central Florida, Doctor of Science, Civil Engineering.

All Calculus 1 Resources

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

Popular Subjects

ACT Tutors in Dallas Fort Worth, GRE Tutors in San Francisco-Bay Area, ISEE Tutors in Miami, Calculus Tutors in Philadelphia, GRE Tutors in Seattle, Chemistry Tutors in Phoenix, Statistics Tutors in Denver, Spanish Tutors in Philadelphia, Spanish Tutors in Washington DC, Math Tutors in Houston

Popular Courses & Classes

SAT Courses & Classes in San Diego, SSAT Courses & Classes in Los Angeles, SAT Courses & Classes in Houston, GMAT Courses & Classes in Dallas Fort Worth, Spanish Courses & Classes in Chicago, ACT Courses & Classes in New York City, GRE Courses & Classes in Denver, Spanish Courses & Classes in Dallas Fort Worth, LSAT Courses & Classes in Washington DC, ACT Courses & Classes in San Diego

Popular Test Prep

MCAT Test Prep in Los Angeles, ACT Test Prep in New York City, LSAT Test Prep in Dallas Fort Worth, GRE Test Prep in Philadelphia, ISEE Test Prep in Washington DC, GRE Test Prep in Los Angeles, SAT Test Prep in Chicago, LSAT Test Prep in San Diego, ACT Test Prep in Los Angeles, 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 : Velocity

Study concepts, example questions & explanations for Calculus 1

All Calculus 1 Resources

Example Questions

Example Question #81 : Spatial Calculus

Example Question #82 : Spatial Calculus

Example Question #83 : Spatial Calculus

Example Question #84 : Spatial Calculus

Example Question #85 : Calculus

Example Question #85 : Spatial Calculus

Example Question #81 : Spatial Calculus

Example Question #81 : Velocity

Example Question #86 : Spatial Calculus

Example Question #87 : Spatial Calculus

All Calculus 1 Resources

Report an issue with this question

DMCA Complaint

Contact Information

Address

Complaint Details

Email address:
Your name:
Feedback: