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AP®MathsCollege BoardCalculus ABStudy GuidesUnit 8: Applications of IntegrationDefinite Integrals in ContextPosition, Velocity & Acceleration

Position, Velocity & Acceleration (College Board AP® Calculus AB)

Study Guide

Jamie Wood

Written by: Jamie Wood

Reviewed by: Dan Finlay

Velocity as an integral

How is velocity defined as an integral?

  • Velocity is the integral of acceleration with respect to time

    • v open parentheses t close parentheses equals integral a open parentheses t close parentheses space d t

      • This follows because acceleration is the derivative of velocity, fraction numerator d over denominator d t end fraction v open parentheses t close parentheses equals a open parentheses t close parentheses

      • and differentiation and integration are inverse operations

  • The definite integral integral subscript t subscript 1 end subscript superscript t subscript 2 end superscript space a open parentheses t close parentheses space italic d t represents the total change in velocity between t equals t subscript 1 and t equals t subscript 2

    • Acceleration is the rate of change of velocity

      • Therefore a open parentheses t close parentheses times straight capital delta t is the change in velocity over a small time interval straight capital delta t

      • a open parentheses t close parentheses space d t is the limit of this change as increment t rightwards arrow 0

      • The integral integral subscript t subscript 1 end subscript superscript t subscript 2 end superscript space a open parentheses t close parentheses space d t sums up all these infinitesimal changes between t equals t subscript 1 and t equals t subscript 2

    • You may also think of this as the integral calculating the area under an acceleration-time graph

      • The change in velocity is equal to this area

  • To find the velocity at a particular point in time, you need to find

    • the change in velocity between times t equals t subscript 1 and t equals t subscript 2

    • then add this on to the velocity at time t equals t subscript 1

      • v open parentheses t subscript 2 close parentheses equals v open parentheses t subscript 1 close parentheses plus integral subscript t subscript 1 end subscript superscript t subscript 2 end superscript space a open parentheses t close parentheses space d t

  • To find an expression for the velocity at any point in time, you need to find

    • the change in velocity between a time t equals t subscript 0 and any other time t

    • then add this on to the velocity at t equals t subscript 0

      • v open parentheses t close parentheses equals v open parentheses t subscript 0 close parentheses plus integral subscript t subscript 0 end subscript superscript t space a open parentheses w close parentheses space d w

      • w here is simply a dummy variable used for the integration

  • Alternatively, find the indefinite integral integral a open parentheses t close parentheses space d t

    • This produces an expression for the velocity, including a constant of integration, plus C

    • Use information in the question about the velocity at a particular point in time to find the value of C

    • This also gives you an expression describing the velocity at any point in time

Examiner Tips and Tricks

Remember that integral subscript t subscript 1 end subscript superscript t subscript 2 end superscript space a open parentheses t close parentheses space italic d t represents the total change in velocity over a period of time, not the final velocity - this is a common error!

Worked Example

The acceleration of a particle for 0 less or equal than t less or equal than 60 seconds is given by the function a defined by a open parentheses t close parentheses equals 1 fourth square root of t, where a open parentheses t close parentheses is measured in meters per second squared.

(a) Find the total change in velocity of the particle between t equals 10 and t equals 20.

Answer:

To find a change in velocity, we can use a definite integral of the acceleration

integral subscript 10 superscript 20 space 1 fourth square root of t d t

Evaluate the integral; factoring out the constant can help

For a question like this, it is likely you could use your calculator to find this integral

table row cell integral subscript 10 superscript 20 space 1 fourth square root of t d t space end cell equals cell space 1 fourth integral subscript 10 superscript 20 space t to the power of 1 half end exponent d t end cell row blank equals cell 1 fourth open square brackets 2 over 3 t to the power of 3 over 2 end exponent close square brackets subscript 10 superscript 20 end cell row blank equals cell 1 fourth open parentheses open square brackets 2 over 3 open parentheses 20 close parentheses to the power of 3 over 2 end exponent close square brackets minus open square brackets 2 over 3 open parentheses 10 close parentheses to the power of 3 over 2 end exponent close square brackets close parentheses end cell row blank equals cell 9.636657... end cell end table

Round the answer to 3 decimal places and state appropriate units

The question asks for a change so you should state if it is an increase or decrease

Increase of 9.637 meters per second

(b) Given that the particle has a velocity of 2 meters per second at time t equals 3 seconds, find the velocity of the particle at t equals 45 seconds.

Answer:

We need to find the change in velocity between t equals 3 and t equals 45, and add this on to the "starting" velocity of 2 meters per second at t equals 3

For a question like this, it is likely you could use your calculator to find this integral

Change in velocity:

table row cell integral subscript 3 superscript 45 space 1 fourth square root of t d t space end cell equals cell space 1 fourth integral subscript 3 superscript 45 space t to the power of 1 half end exponent d t end cell row blank equals cell 1 fourth open square brackets 2 over 3 t to the power of 3 over 2 end exponent close square brackets subscript 3 superscript 45 end cell row blank equals cell 1 fourth open parentheses 2 over 3 open parentheses 45 close parentheses to the power of 3 over 2 end exponent minus open parentheses 2 over 3 open parentheses 3 close parentheses to the power of 3 over 2 end exponent close parentheses close parentheses end cell row blank equals cell 49.445504... end cell end table

This is the change in velocity between t equals 3 and t equals 45, so we need to add on the starting value at t equals 3, which is 2 meters per second

2 plus 49.445504... equals 51.445507...

Round the answer to 3 decimal places and state appropriate units

51.446 meters per second

Position as an integral

How is position defined as an integral?

  • Position, or displacement, is the integral of velocity with respect to time

    • s open parentheses t close parentheses equals integral v open parentheses t close parentheses space d t

      • This follows because velocity is the derivative of displacement, fraction numerator d over denominator d t end fraction s open parentheses t close parentheses equals v open parentheses t close parentheses

      • and differentiation and integration are inverse operations

  • The definite integral integral subscript t subscript 1 end subscript superscript t subscript 2 end superscript space v open parentheses t close parentheses space italic d t represents the total change in displacement between t equals t subscript 1 and t equals t subscript 2

    • Velocity is the rate of change of displacement

      • Therefore v open parentheses t close parentheses times straight capital delta t is the change in velocity over a small time interval straight capital delta t

      • v open parentheses t close parentheses space d t is the limit of this change as increment t rightwards arrow 0

      • The integral integral subscript t subscript 1 end subscript superscript t subscript 2 end superscript space v open parentheses t close parentheses space d t sums up all these infinitesimal changes between t equals t subscript 1 and t equals t subscript 2

    • You may also think of this as the integral calculating the area under a velocity-time graph

      • The change in displacement is equal to this area

  • To find the displacement at a particular point in time, you need to find

    • the change in displacement between times t equals t subscript 1 and t equals t subscript 2

    • then add this on to the displacement at time t equals t subscript 1

      • s open parentheses t subscript 2 close parentheses equals s open parentheses t subscript 1 close parentheses plus integral subscript t subscript 1 end subscript superscript t subscript 2 end superscript space v open parentheses t close parentheses space d t

  • To find an expression for the displacement at any point in time, you need to find

    • the change in displacement between a time t equals t subscript 0 and any other time t

    • then add this on to the displacement at t equals t subscript 0

      • s open parentheses t close parentheses equals s open parentheses t subscript 0 close parentheses plus integral subscript t subscript 0 end subscript superscript t space v open parentheses w close parentheses space d w

      • w here is simply a dummy variable used for the integration

  • Alternatively, find the indefinite integral integral v open parentheses t close parentheses space d t

    • This produces an expression for the displacement, including a constant of integration, plus C

    • Use information in the question about the displacement or position at a particular point in time to find the value of C

    • This also gives you an expression describing the displacement at any point in time

Examiner Tips and Tricks

Remember that integral subscript t subscript 1 end subscript superscript t subscript 2 end superscript space v open parentheses t close parentheses space italic d t represents the total change in displacement over a period of time

  • It is not the final displacement or position

  • It is also not the total distance traveled

  • Both of these are common errors!

Worked Example

A particle moves along the x-axis with a velocity described by the function

v open parentheses t close parentheses equals 12 t minus 1 over 20 t cubed for 0 less or equal than t less or equal than 24

v is measured in feet per second and t is measured in seconds.

(a) Given that at time t equals 3 the particle is at a displacement of 40 feet from the origin, find the displacement of the particle from the origin at time t equals 10.

Answer:

We need to find the total change in displacement from t equals 3 to t equals 10, and add this on to 40 feet, which was the displacement at t equals 3

The total change in displacement is found by integrating the velocity

For a question like this, it is likely you could use your calculator to find this integral

table row cell integral subscript 3 superscript 10 space 12 t minus 1 over 20 t cubed space italic d t end cell equals cell open square brackets 6 t squared minus 1 over 80 t to the power of 4 close square brackets subscript 3 superscript 10 end cell row blank equals cell 6 open parentheses 10 close parentheses squared minus 1 over 80 open parentheses 10 to the power of 4 close parentheses minus open parentheses 6 open parentheses 3 close parentheses squared minus 1 over 80 open parentheses 3 to the power of 4 close parentheses close parentheses end cell row blank equals cell 422.0125 end cell end table

Add this on to the displacement at t equals 3 and round to 3 decimal places

422.0125 plus 40 equals 462.0125

At t equals 10, the particle will be 462.013 feet from the origin

(b) The particle starts it's motion at t equals 0, with a displacement of zero feet.

Find the length of time it takes for the particle to return the same position that it started in.

Answer:

When the particle is back in the same place it started, its displacement will be zero again

You could think of this as being a total change of zero from the starting point

Because s open parentheses 0 close parentheses equals 0, the definite integral of the velocity starting at t equals 0 will be equal to the total displacement

Use this fact, and solve for an unknown upper limit, T

s equals integral subscript 0 superscript T 12 t minus 1 over 20 t cubed space d t space equals space 0

Integrate and substitute in the limits

table row cell open square brackets 6 t squared minus 1 over 80 t to the power of 4 close square brackets subscript 0 superscript T end cell equals 0 row cell open square brackets 6 T squared minus 1 over 80 T to the power of 4 close square brackets minus open square brackets 0 minus 0 close square brackets end cell equals 0 row cell 6 T squared minus 1 over 80 T to the power of 4 end cell equals 0 end table

Factorize and solve

T squared open parentheses 6 minus 1 over 80 T squared close parentheses equals 0

T squared equals 0 or 6 minus 1 over 80 T squared equals 0

T equals 0 or T equals square root of 480 equals 4 square root of 30 equals 21.908902...

T equals 0 corresponds to the start of the motion when the displacement was also 0

Round the answer to 3 decimal places

It takes 21.909 seconds for the particle to return to its starting place

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    Author
    Reviewer
    Jamie Wood

    Author: Jamie Wood

    Expertise: Maths

    Jamie graduated in 2014 from the University of Bristol with a degree in Electronic and Communications Engineering. He has worked as a teacher for 8 years, in secondary schools and in further education; teaching GCSE and A Level. He is passionate about helping students fulfil their potential through easy-to-use resources and high-quality questions and solutions.

    Dan Finlay

    Author: Dan Finlay

    Expertise: Maths Lead

    Dan graduated from the University of Oxford with a First class degree in mathematics. As well as teaching maths for over 8 years, Dan has marked a range of exams for Edexcel, tutored students and taught A Level Accounting. Dan has a keen interest in statistics and probability and their real-life applications.