# Question #089b1

Dec 22, 2016

Kinematic equation of interest is

$v \left(t\right) = u + a t$ .....(1)
where $v \left(t\right)$ is velocity after time $t$, $u$ is initial velocity of an object and $a$ is constant acceleration experienced by it.

1. Recall the expression
$\text{Displacement"="Velocity"xx"time}$
2. Observe it looks like equation of a straight line in the form
$y = m x + c$.

We know that velocity is rate of change of displacement, therefore equation (1) can be written as

$\frac{\mathrm{ds} \left(t\right)}{\mathrm{dt}} = u + a t$
$\implies \mathrm{ds} \left(t\right) = \left(u + a t\right) \cdot \mathrm{dt}$ .....(2)

If we integrate both sides we get
$\int \mathrm{ds} \left(t\right) = {\int}_{{t}_{0}}^{t} \left(u + a t\right) \cdot \mathrm{dt}$
$\implies s \left(t\right) = {\int}_{{t}_{0}}^{t} \left(u + a t\right) \cdot \mathrm{dt}$ ......(3)
We see that LHS of the equation is total displacement, and RHS is area under the velocity-time graph from time ${t}_{0}$ to $t$.
Equation (3) is the required expression.

One should not be surprised if one calculates integral of RHS of equation (3) from time $t = 0$ to $t$, one actually obtains the other kinematic equation

$s = u t + \frac{1}{2} a {t}^{2}$