Higher Engineering Mathematics, Sixth Edition

Some applications of differentiation 301

x

t

Time

Distance

Figure 28.1

x

t

B

A

Time

Distance

Figure 28.2

for the distancexis known in terms of timetthen the
velocity is obtained by differentiating the expression.
The accelerationaof the car is defined as the rate
of change of velocity. A velocity/time graph is shown
in Fig. 28.3. If δvis the change inv andδt the

corresponding change in time, thena=

δv

δt

.
Asδt→0, the chordCDbecomes a tangent, such that
at pointC, the acceleration is given by:

a=

dv

dt

Hence the acceleration of the car at any instant is
given by the gradient of the velocity/time graph. If an
expression for velocity is known in terms of timet
then the acceleration is obtained by differentiating the
expression.

Acceleration a=

dv

dt

.However,v=

dx

dt

. Hence

a=

d

dt

(

dx

dt

)

=

d^2 x

dx^2

v

t

D

C

Time

Velocity

Figure 28.3

The acceleration is given by the second differential

coefficient of distancexwith respect to timet.

Summarizing, if a body moves a distancexmetres

in a timetseconds then:

(i) distancex=f(t).

(ii) velocityv=f′(t)or

dx

dt

, which is the gradient of

the distance/time graph.

(iii) accelerationa=

dv

dt

=f′′(t)or

d^2 x

dt^2

, which is the

gradient of the velocity/time graph.

Problem 5. The distancexmetres moved

by a car in a timetseconds is given by

x= 3 t^3 − 2 t^2 + 4 t−1. Determine the velocity and

acceleration when (a)t=0and(b)t= 1 .5s.

Distance x= 3 t^3 − 2 t^2 + 4 t−1m

Velocity v=

dx

dt

= 9 t^2 − 4 t+4m/s

Acceleration a=

d^2 x

dx^2

= 18 t−4m/s^2

(a) When timet=0,

velocityv= 9 ( 0 )^2 − 4 ( 0 )+ 4 =4m/sand

acceleration a= 18 ( 0 )− 4 =−4m/s^2 (i.e. a

deceleration)

(b) When timet= 1 .5s,

velocityv= 9 ( 1. 5 )^2 − 4 ( 1. 5 )+ 4 =18.25m/s

and accelerationa= 18 ( 1. 5 )− 4 =23m/s^2