We are open Saturday and Sunday!

Call Now to Set Up Tutoring:

(888) 888-0446

Calculus 1 : How to find 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

← Previous 1 2 … 5 6 7 8 9 10 11 12 13 … 17 18 Next →

Example Question #931 : Spatial Calculus

Given that the final acceleration is $32 \frac{m}{s^2}$ , and that the final position is 44m with a velocity of $45\frac{m}{s}$ after 10 seconds, find the initial position at t=0 .

Possible Answers:

-1564m

1194m

1040m

Correct answer:

1194m

Explanation:

We have to start from acceleration and integrate back to position to determine the initial position at t=0 .

$V(t)=\int 32= 32t+V_0$

Since we know that the velocity at t=10 is $45\frac{m}{s}$ , we plug it in to determine the initial velocity

V(10)=45=32(10)+V_0 , V_0=-275

V(t)=32t-275

Now that we know the function for velocity, we can integrate it to find position

$P(t)= \int V(t)=16t^2-275t+P_0$

Since we also know that at t=10 that the position is at 44m , we use it to find initial position at t=0

P(10)=44=16(10)^2-275(10)+P_0 , P_0= 1194

Therefore, at t=0, the initial position is 1194m.

Report an Error

Example Question #932 : Spatial Calculus

Find the position equation of the particle, given the following information:

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

s(0)=0

Possible Answers:

$\sin(t)+t^2$

$\sin(t)+\frac{t^2}{2}$

$\cos(t)+t^2+1$

$-\sin(t)+t^2$

Correct answer:

$\sin(t)+t^2$

Explanation:

In order to find the position equation, we must integrate the velocity function and then plug in the initial condition to solve for C:

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

We used the following rules for integration:

$\int\cos(x)dx=\sin(x)+C$ , $\int x^ndx=\frac{1}{n+1}x^{n+1}+C$

Now, plug in t=0 into the position equation and solve for C:

s(0)=0=0+0+C

Thus, C=0, and our final answer is

$\sin(t)+t^2$

Report an Error

Example Question #933 : Spatial Calculus

The velocity of a particle is given by the function v(t)=2e^t-3sin(t) .

If the particle has an initial position of , what is its position at time $t=\pi$ ?

Possible Answers:

$2e^{\pi}-5$

$2e^{\pi}-3$

$2e^{\pi}$

$2e^{\pi}+2$

$2e^{\pi}-8$

Correct answer:

$2e^{\pi}-8$

Explanation:

The position function can be found by integrating the velocity function with respect to time:

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

Velocity is given as:

v(t)=2e^t-3sin(t)

So the position function is:

x(t)=2e^t+3cos(t)+C

The integration of constant is found by using the initial condition provided:

x(0)=0=2e^0+3cos(0)+C

C=-5

Therefore

x(t)=2e^t+3cos(t)-5

$x(\pi)=2e^{\pi}-3-5$

$x(\pi)=2e^{\pi}-8$

Report an Error

Example Question #934 : Spatial Calculus

The velocity of a particle is given by the function $v(t)=6t+e^{-3t}$ . If the particle has an initial position of , what will its position be at time t=2 ?

Possible Answers:

$\frac{37-e^{-6}}{3}$

$\frac{35+e^{-6}}{3}$

$12+\frac{e^{-6}}{3}$

$12-\frac{e^{-6}}{3}$

Correct answer:

$\frac{37-e^{-6}}{3}$

Explanation:

Position can be found by integrating velocity with respect to time:

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

For the velocity function

$v(t)=6t+e^{-3t}$

The position function is:

$x(t)=3t^2-\frac{e^{-3t}}{3}+C$

The constant of integration can be found by using the initial condition:

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

$C=\frac{1}{3}$

So the definite integral is:

$x(t)=3t^2-\frac{e^{-3t}}{3}+\frac{1}{3}$

$x(2)=12-\frac{e^{-6}}{3}+\frac{1}{3}$

$x(2)=\frac{37-e^{-6}}{3}$

Report an Error

Example Question #935 : Spatial Calculus

Find the position function of a ball thrown by a person tall if it's initial velocity is $v_{0}=40 m/s$ and its acceleration is $a=-9.8m/s^{2}$ .

Possible Answers:

$s(t)=2+40t-9.8t^{2}$

s(t)=2

$s(t)=40t-9.8t^{2}$

$s(t)=2+40t-4.9t^{2}$

s(t)=2+40t

Correct answer:

$s(t)=2+40t-4.9t^{2}$

Explanation:

The position function of an object moving with uniform acceleration is $s(t)=s_{0}+v_{0}t+\frac{1}{2}at^{2}$ , where $s_{0}$ is the inital position of the object, $v_{0}$ is the inital velocity of the object and is the acceleration of the object.

For this problem:

$s_{0}=2 m$ , $v_{0}=40m/s$ , $a=-9.8m/s^{2}$

$s(t)=2+40t+\frac{1}{2}(-9.8)t^{2}$

$s(t)=2+40t-4.9t^{2}$

Report an Error

Example Question #936 : Spatial Calculus

Find the position of a ball after seconds if it is thrown by a person tall and its initial velocity is $v_{0}=40 m/s$ and its acceleration is $a=-9.8m/s^{2}$ .

Possible Answers:

s=79.5

s=5

s=40

s=2

s=-9.8

Correct answer:

s=79.5

Explanation:

To find the position of the object after a certain time, we first must find the position function of the object, then solve the position function at that given time.

For this problem:

$s_{0}=2 m$ , $v_{0}=40m/s$ , $a=-9.8m/s^{2}$ , t=5s

$s(t)=2+40t+\frac{1}{2}(-9.8)t^{2}$

$s(t)=2+40t-4.9t^{2}$

$s(5)=2+40(5)-4.9(5)^{2}=2+200-122.5=79.5$

After seconds, the ball has travelled 79.5 meters.

Report an Error

Example Question #941 : Spatial Calculus

Find the position function of a rocket shot from the ground if its initial velocity is $v_{0}=100ft/s$ and its acceleration is $a=-32ft/s^{2}$ .

Possible Answers:

$s(t)=100t-16t^{2}$

s(t)=-32t+100

s(t)=100

$s(t)=t^{2}+5t-12$

$s(t)=100t-32t^{2}$

Correct answer:

$s(t)=100t-16t^{2}$

Explanation:

For this problem:

$s_{0}=0 ft$ , $v_{0}=100ft/s$ , $a=-32ft/s^{2}$

$s(t)=100t+\frac{1}{2}(-32)t^{2}$

$s(t)=100t-16t^{2}$

Report an Error

Example Question #82 : How To Find Position

A ball is thrown straight upward from ground level. It has an initial velocity of 20m/s . At what time will the ball reach its maximum height?

Use to approximate acceleration due to gravity.

Possible Answers:

t=1

t=5

t=0

t=4

t=2

Correct answer:

t=2

Explanation:

The ball's acceleration will simply be its acceleration due to gravity, approximated with .

a(t)=-10

Integrating the acceleration function will leave you with the function for velocity.

v(t) = -10t + C

You know that has an initial velocity of 20m/s ., so v(0)=20 . Use this to find the value of the integration constant, .

20 = -10(0) + C

20=C

Therefore, our velocity funciton is v(t) = -10t +20

Finding the zero of the velocity function will give local extrema of the position function.

0= -10t +20

-20=-10t

2=t

Since the velocity of the ball changes from positive to negative about t=2 , this is a local maximum. Therefore, the ball will reach its maximum height when t=2 .

Report an Error

Example Question #83 : How To Find Position

A vehicle at a position of 5 m . It accelerates from rest according to the acceleration function, a(t)

a(t)= 1.5t+4

where is time (in seconds).

Find the function representing the vehicle's position.

Possible Answers:

x(t)=0.25t^3+2t^2+5

x(t)=9t^2+24t+5

x(t)=9t^3+24t^2+5

x(t)=0.25t^3+2t^2

x(t)=9t^3+24t^2

Correct answer:

x(t)=0.25t^3+2t^2+5

Explanation:

Integrating the acceleration function will result in the function representing the vehicle's velocity. To integrate this function use the rule

$x^n\rightarrow \frac{x^{n+1}}{n+1}$ .

Therefore,

a(t)= 1.5t+4

v(t) = 0.75t^2 + 4t + C .

You know that the vehicle begins at rest, so v(0)=0 . Use this to find the value of the integration constant, .

0 = 0.75(0)^2 + 4(0) + C

0=C

Therefore, our velocity funciton is v(t) = 0.75t^2 + 4t

Integrating the velocity function will result in the function representing the vehicle's position.

v(t) = 0.75t^2 + 4t

x(t)=0.25t^3+2x^2+C'

You know that the vehicle begins at a position of 5 m , so x(0)=5 .

Use this to find the value of the integration constant.

5=0.25(0^3)+2(0^2)+C'

5=C'

Therefore,

x(t)=0.25t^3+2t^2+5

Report an Error

Example Question #84 : How To Find Position

An ant is moving with a velocity given by the following function:

$v(t)=\sec^2(t)+2t+1$

What is the position function of the ant?

Possible Answers:

$\tan(t)+t^2+t$

$\tan(t)+t^2+1+C$

$\tan(t)+\frac{t^2}{2}+t+C$

$\tan(t)+t^2+t+C$

Correct answer:

$\tan(t)+t^2+t+C$

Explanation:

To find the position function of the ant, we must integrate the velocity function:

$s(t)=\int v(t)=\int (\sec^2(t)+2t+1)dt=\tan(t)+t^2+t+C$

The integration was performed using the following rules:

$\int \sec^2(x)dx=\tan(x)+C$ ,

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

Report an Error

← Previous 1 2 … 5 6 7 8 9 10 11 12 13 … 17 18 Next →

View Calculus Tutors

Karl
Certified Tutor

Tabor College, Bachelor in Arts, Mathematics Teacher Education. MidAmerica Nazarene University, Masters in Education, Educati...

View Calculus Tutors

Anabel
Certified Tutor

UDELAR, Diploma, Chemistry. IPA, Bachelor of Education, Mathematics.

View Calculus Tutors

Anthony
Certified Tutor

George Washington University, Bachelor of Science, Biology, General. Columbia University in the City of New York, Master of S...

All Calculus 1 Resources

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

Popular Subjects

Computer Science Tutors in Seattle, English Tutors in Philadelphia, Math Tutors in Philadelphia, Spanish Tutors in Dallas Fort Worth, MCAT Tutors in Seattle, Computer Science Tutors in Washington DC, Chemistry Tutors in Miami, SSAT Tutors in Boston, Biology Tutors in San Diego, Calculus Tutors in Chicago

Popular Courses & Classes

SAT Courses & Classes in Miami, GMAT Courses & Classes in Phoenix, MCAT Courses & Classes in San Diego, MCAT Courses & Classes in Denver, SAT Courses & Classes in Philadelphia, Spanish Courses & Classes in Dallas Fort Worth, ISEE Courses & Classes in Houston, ACT Courses & Classes in Boston, GMAT Courses & Classes in Denver, ACT Courses & Classes in Houston

Popular Test Prep

SAT Test Prep in New York City, SSAT Test Prep in Philadelphia, ISEE Test Prep in San Francisco-Bay Area, GMAT Test Prep in Seattle, LSAT Test Prep in Los Angeles, GRE Test Prep in Atlanta, SAT Test Prep in Chicago, ACT Test Prep in Boston, ACT Test Prep in Dallas Fort Worth, ISEE Test Prep in Philadelphia

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 position

Study concepts, example questions & explanations for Calculus 1

All Calculus 1 Resources

Example Questions

Example Question #931 : Spatial Calculus

Example Question #932 : Spatial Calculus

Example Question #933 : Spatial Calculus

Example Question #934 : Spatial Calculus

Example Question #935 : Spatial Calculus

Example Question #936 : Spatial Calculus

Example Question #941 : Spatial Calculus

Example Question #82 : How To Find Position

Example Question #83 : How To Find Position

Example Question #84 : How To Find Position

All Calculus 1 Resources

Report an issue with this question

DMCA Complaint

Contact Information

Address

Complaint Details

Email address:
Your name:
Feedback: