Cambridge Additional Mathematics

Applications of differential calculus (Chapter 14) 383

3 A particle moves in a straight line with velocity function v(t)=2

p

t+3cm s¡^1 , t> 0.

a Find the average acceleration from t=1to t=4seconds.

b Find the average acceleration from t=1to t=1+h seconds.

c Find the value of lim

h! 0

v(1 +h)¡v(1)

h

. Interpret this value.

d Interpret lim

h! 0

v(t+h)¡v(t)

h

.

4 An object moves in a straight line with displacement functions(t) and velocity function v(t), t> 0.

State the meaning of:

a lim

h! 0

s(4 +h)¡s(4)

h

b lim

h! 0

v(4 +h)¡v(4)

h

VELOCITY AND ACCELERATION FUNCTIONS

If a particle P moves in a straight line and its position is given by the displacement function s(t), t> 0 ,

then:

² thevelocityof P at timetis given by v(t)=s^0 (t)

² theaccelerationof P at timetis given by a(t)=v^0 (t)=s^00 (t)

² s(0), v(0), and a(0) give us the position, velocity, and acceleration of the

particle at time t=0, and these are called theinitial conditions.

SIGN INTERPRETATION

Suppose a particle P moves in a straight line with displacement function s(t) relative to an origin O. Its

velocity function is v(t) and its acceleration function is a(t).

We can usesign diagramsto interpret:

² where the particle is located relative to O

² the direction of motion and where a change of direction occurs

² when the particle’s velocity is increasing or decreasing.

SIGNS OFs(t):

s(t) Interpretation

=0 PisatO

> 0 P is located to the right of O

< 0 P is located to the left of O

SIGNS OFv(t):

v(t) Interpretation

=0 P is instantaneously at rest

> 0 P is moving to the right

< 0 P is moving to the left

SIGNS OFa(t): a(t) Interpretation

> 0 velocity is increasing

< 0 velocity is decreasing

=0 velocity may be a maximum or minimum or possibly constant

4037 Cambridge

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Y:\HAESE\CAM4037\CamAdd_14\383CamAdd_14.cdr Monday, 7 April 2014 10:28:25 AM BRIAN