AP®MathsCollege BoardCalculus ABStudy GuidesUnit 8: Applications of IntegrationDefinite Integrals in ContextDistance & Speed

Distance & Speed (College Board AP® Calculus AB): Study Guide

Jamie Wood

Written by: Jamie Wood

Reviewed by: Dan Finlay

Updated on

College Board Calculus ABCollege Board Calculus BC

Distance & speed as integrals

How are distance and speed different from displacement and velocity?

  • Distance is the magnitude of displacement, open vertical bar s close vertical bar

    • Travelling 3 meters forwards and 3 meters backwards is a distance of 6 meters traveled

    • But the displacement after this motion would be zero

  • Speed is the magnitude of velocity, open vertical bar v close vertical bar

    • Travelling at 10 meters per second forward is a velocity of +10 meters per second

    • Travelling at 10 meters per second backwards is a velocity of -10 meters per second

    • In both these cases however, the speed is 10 meters per second

How do I find a distance using integration?

  • The total distance traveled between times t subscript 1 and t subscript 2 is given by integral subscript t subscript 1 end subscript superscript t subscript 2 end superscript space open vertical bar v open parentheses t close parentheses close vertical bar space d t

    • This is because even when the object is moving backwards (negative velocity), the distance traveled is increasing (whilst the displacement is decreasing)

  • Consider a simple example, where v equals 2 t minus 6

    • To find the change in displacement between t equals 0 and t equals 6 we would calculate

      • integral subscript 0 superscript 6 space 2 t minus 6 space italic d t, which has a value of zero

      • This is because integral subscript 0 superscript 3 space 2 t minus 6 space italic d t equals negative 9 and integral subscript 3 superscript 6 space 2 t minus 6 space italic d t equals plus 9

    • To find the distance traveled between t equals 0 and t equals 6 we would calculate

      • integral subscript 0 superscript 6 space open vertical bar 2 t minus 6 close vertical bar space italic d t, which has a value of 18

      • You could find this on your calculator directly, using the "absolute" (i.e. absolute value or modulus) function

      • Alternatively you can sketch a velocity-time graph to find where the area under the graph would be negative and positive, and split the calculation into two integrals

      • Then make any negative integrals positive

        • integral subscript 0 superscript 6 space open vertical bar 2 t minus 6 close vertical bar space italic d t equals open vertical bar integral subscript 0 superscript 3 space 2 t minus 6 space italic d t close vertical bar space plus space integral subscript 3 superscript 6 space 2 t minus 6 space italic d t equals open vertical bar negative 9 close vertical bar space plus space 9 space equals 18

Graph of velocity (v) versus time (t) with shaded areas representing integrals from 0 to 3 and 3 to 6. Text: Change in displacement is 0, Distance traveled is 18.

How does using speed rather than velocity affect calculations?

  • Pay close attention to whether a question refers to speed or velocity

  • This is important when a calculation involving velocity "goes through zero"

  • E.g. If a particle with velocity 5 meters per second decreases its velocity by 20 meters per second

    • Its new velocity is -15 meters per second

    • However its new speed is 15 meters per second

  • If a particle with velocity -8 meters per second, increases its velocity by 10 meters per second

    • Its new velocity is 2 meters per second

    • Its new speed is also 2 meters per second

    • But, it has changed from a speed of 8 meters per second, to a speed of 2 meters per second

      • So it is true to say its speed has decreased by 6 meters per second

  • To find a change in velocity, leading to a calculation similar to above, you may need to integrate an expression for acceleration

Worked Example

The acceleration of a particle over the interval 0 less or equal than t less or equal than 6 pi is described by the function a open parentheses t close parentheses equals 3 over 2 cos space open parentheses t over 2 close parentheses where v is measured in feet per second and t is measured in seconds.

At t equals 0, the particle is at rest.

(a) Calculate the change in speed between t equals fraction numerator 3 pi over denominator 2 end fraction and t equals 3 pi. State if this is an increase or decrease.

Answer

Start by finding an expression for the velocity by integrating the expression for the acceleration

table row cell v open parentheses t close parentheses end cell equals cell integral a open parentheses t close parentheses space d t space equals space integral 3 over 2 cos open parentheses t over 2 close parentheses space d t end cell row cell v open parentheses t close parentheses end cell equals cell space 3 sin t over 2 space plus space C end cell end table

To find the constant of integration, substitute in a known velocity at a point in time

We were told in the question the initial velocity is zero, so v equals 0 when t equals 0

table row 0 equals cell 3 sin open parentheses 0 close parentheses plus C end cell row 0 equals cell 0 plus C end cell row C equals 0 end table

table row cell v open parentheses t close parentheses end cell equals cell 3 sin t over 2 end cell end table

Find the velocities at t equals fraction numerator 3 pi over denominator 2 end fraction and t equals 3 pi

table row cell v open parentheses fraction numerator 3 pi over denominator 2 end fraction close parentheses end cell equals cell 3 sin fraction numerator 3 pi over denominator 4 end fraction equals 2.12132034... end cell end table

table row cell v open parentheses 3 pi close parentheses end cell equals cell 3 sin fraction numerator 3 pi over denominator 2 end fraction equals negative 3 end cell end table

Consider these results in terms of speeds rather than velocities

Speed at t equals fraction numerator 3 pi over denominator 2 end fraction is 2.121320344...

Speed at t equals 3 pi is 3

Therefore the speed has increased

Increase in speed = 3 minus 2.12132034... space equals space 0.878679656...

Round to 3 decimal places

The speed has increased by 0.879 feet per second

(b) Find the total distance traveled by the particle between t equals straight pi and t equals 4 straight pi.

Answer:

Distance traveled will be integral subscript pi superscript 4 pi end superscript space open vertical bar v open parentheses t close parentheses close vertical bar space d t equals integral subscript pi superscript 4 pi end superscript space open vertical bar 3 sin t over 2 close vertical bar space d t

On a calculator question you could use your calculator to evaluate this

If doing it by hand, you have to be careful with negative areas (i.e. intervals where v open parentheses t close parentheses less than 0)

The graph of v equals 3 sin space t over 2 will intersect the horizontal axis at 0 comma space 2 pi comma space 4 pi comma space 6 pi comma space..., so the area between pi and 4 pi will have a positive and a negative portion

Split the integral into multiple parts, with the horizontal axis intercepts as the boundaries

table row cell integral subscript pi superscript 2 pi end superscript space 3 sin space t over 2 d t space end cell equals cell open square brackets negative 6 cos t over 2 close square brackets subscript pi superscript 2 pi end superscript end cell row blank equals cell space minus 6 cos pi minus open parentheses negative 6 cos pi over 2 close parentheses end cell row blank equals cell negative 6 open parentheses negative 1 close parentheses minus open parentheses 0 close parentheses end cell row blank equals 6 end table

table row cell integral subscript 2 pi end subscript superscript 4 pi end superscript space 3 sin space t over 2 d t space end cell equals cell open square brackets negative 6 cos t over 2 close square brackets subscript 2 pi end subscript superscript 4 pi end superscript end cell row blank equals cell space minus 6 cos 2 pi minus open parentheses negative 6 cos pi close parentheses end cell row blank equals cell negative 6 open parentheses 1 close parentheses minus open parentheses negative 6 open parentheses negative 1 close parentheses close parentheses end cell row blank equals cell negative 6 minus 6 end cell row blank equals cell negative 12 end cell end table

Find the total distance travelled by considering the absolute values of the changes in displacement

I.e. distance space traveled equals integral subscript pi superscript 4 pi end superscript space open vertical bar 3 sin t over 2 close vertical bar space d t equals integral subscript pi superscript 2 pi end superscript space 3 sin t over 2 space d t plus open vertical bar integral subscript 2 pi end subscript superscript 4 pi end superscript space 3 sin t over 2 space d t close vertical bar

6 + 12 = 18

Distance of 18 feet traveled between t equals pi and t equals 4 pi

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    Author
    Reviewer
    Jamie Wood

    Author: Jamie Wood

    Expertise: Maths Content Creator

    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

    Reviewer: Dan Finlay

    Expertise: Maths Subject 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.