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

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