Cambridge Additional Mathematics

446 Applications of integration (Chapter 16)

DISPLACEMENT AND VELOCITY FUNCTIONS

In this section we are concerned withmotion in a straight line.

For some displacement function s(t), the velocity function is v(t)=s^0 (t).

So, given a velocity function we can determine the displacement function by the integral

s(t)=

Z

v(t)dt

The constant of integration determines theinitial positionon the line where the object begins.

Using the displacement function we can determine the change in displacement in a time intervalt 16 t 6 t 2

using the integral:

Displacement=s(t 2 )¡s(t 1 )=

Zt 2

t 1

v(t)dt

TOTAL DISTANCE TRAVELLED

To determine the total distance travelled in a time interval t 16 t 6 t 2 , we need to account for any changes

of direction in the motion.

To find the total distance travelled given a velocity function v(t)=s^0 (t) on t 16 t 6 t 2 :

² Draw a sign diagram for v(t) so we can determine any changes of direction.

² Determine s(t) by integration, including a constant of integration.

² Find s(t 1 ) and s(t 2 ). Also find s(t) at each time the direction changes.

² Draw a motion diagram.

² Determine the total distance travelled from the motion diagram.

VELOCITY AND ACCELERATION FUNCTIONS

We know that the acceleration function is the derivative of the velocity function, so a(t)=v^0 (t).

So, given an acceleration function, we can determine the velocity function by integration:

v(t)=

Z

a(t)dt

Summary

s(t)

displacement

v(t)=

ds

dt

velocity

a(t)=

dv

dt

=

d^2 s

dt^2

acceleration

differentiate differentiate

integrate integrate

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Additional Mathematics
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