300 DIFFERENTIAL CALCULUS
Figure 28.2
Hence the velocity of the car at any instant is given by
the gradient of the distance/time graph. If an expres-
sion for the distancexis known in terms of timet
then 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. Ifδvis the change invandδtthe
corresponding change in time, thena=
δv
δt.Figure 28.3
Asδt→0, the chordCDbecomes a tangent, such
that at pointC, the acceleration is given by:
a=dv
dtHence 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 time
tthen 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
The acceleration is given by the second dif-
ferential coefficient of distancexwith respect to
timet.
Summarising, if a body moves a distance
xmetres in a timetseconds then:(i)distancex=f(t).(ii)velocityv=f′(t)ordx
dt, which is the gradient
of the distance/time graph.(iii)accelerationa=dv
dt=f′′(t)ord^2 x
dt^2, which is
the gradient of the velocity/time graph.Problem 5. The distance x metres moved
by a car in a time t seconds is given by
x= 3 t^3 − 2 t^2 + 4 t−1. Determine the velocity
and acceleration when (a)t=0 and (b)t= 1 .5s.Distance x= 3 t^3 − 2 t^2 + 4 t−1mVelocity v=dx
dt= 9 t^2 − 4 t+4m/sAcceleration 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 =18.25 m/s
and accelerationa=18(1.5)− 4 =23 m/s^2Problem 6. Supplies are dropped from a heli-
coptor and the distance fallen in a timetseconds
is given byx=^12 gt^2 , whereg= 9 .8 m/s^2. Deter-
mine the velocity and acceleration of the sup-
plies after it has fallen for 2 seconds.