### Applications of differentiation Part VI

Displacement, Velocity and Acceleration

Consider the motion of a particle which moves in a way such that its displacement s from a fixed point O at time t seconds is given by

$s = -t^2 + 5t + 6$

The instantaneous velocity is given by the differentiation of displacement

$v = \frac{ds}{dt} = -2t + 5$

The instantaneous acceleration is given by the differentiation of velocity

$a = \frac{dv}{dt} = -2$

To obtain velocity function from accleration, we integrate the acceleration function

To obtain the displacement function from velocity, we integrate the velocity function

Example 1

The acceleration of a particle is given by the function $a = 10 + 15t^2$ where t is the time in seconds after leaving start point O. Find the velocity of the particle at t = 1 given that the initial velocity of the particle is 0m/s.

To obtain velocity function from accleration, we integrate the acceleration function

\begin{aligned} v &= \int a \:dt \\ &= \int 10 + 15t^2 \: dt \\ &= 10 t + \frac{15t^3}{3} + c\\ &= 10 t + 5t^3 + c \\ \text{When t=0, v=0, therefore c = 0} \\ v&= 10 t + 5t^3 \\ \text{Substitute t =1 } v&=10+5 \\ &=15 \\ \end{aligned}

What's Next ?

Check out the Basic Techniques of Differentiation
Check out the Contents Page