Cambridge Additional Mathematics

444 Applications of integration (Chapter 16)

DISTANCES FROM VELOCITY GRAPHS

Suppose a car travels at a constant positive velocity of 60 km h¡^1 for 15 minutes.

We know the distance travelled=speed£time

=60km h¡^1 £^14 h

=15km.

When we graphvelocityagainsttime, the graph is a horizontal

line, and we can see that the distance travelled is the area

shaded.

So, the distance travelled can also be found by the definite

integral

Z^1

4

0

60 dt=15km.

Now suppose the velocity decreases at a constant rate, so that

the car, initially travelling at 60 km h¡^1 , stops in 6 minutes or

1

10 hour.

In this case theaveragespeed is 30 km h¡^1 , so the distance

travelled =30km h¡^1 £ 101 h

=3km

But the triangle has area=^12 £base£altitude

=^12 £ 101 £60 = 3

So, once again the shaded area gives us the distance travelled, and we can find it using the definite integral

Z^1

10

0

(60¡ 600 t)dt=3.

These results suggest that:

For a velocity-time function v(t) where v(t)> 0 on the

interval t 16 t 6 t 2 ,

distance travelled=

Zt 2

t 1

v(t)dt.

If we have a change of direction within the time interval then the velocity will change sign. We therefore

need to add the components of area above and below thet-axis to find the total distance travelled.

C Kinematics

60

Qr time(thours)

v(t) = 60¡¡

O

velocity km h()-1

60

O

velocity km h()-1

time

(thours)

Aq_p

v(t) = 60 - 600t

v

O t^1 t^2 t

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Additional Mathematics
Y:\HAESE\CAM4037\CamAdd_16\444CamAdd_16.cdr Monday, 7 April 2014 4:17:57 PM BRIAN