Derive the equations of uniformly accelerated motion by the graphical method.

(N/A) Let the velocity of the particle at $t=0$ be $v_{0}$ and at time $t$ be $v$.
$1$. First Equation of Motion $(v = v_{0} + at)$:
The acceleration $a$ is the slope of the velocity-time graph line $AB$.
$a = \text{slope} = \frac{v - v_{0}}{t - 0} = \frac{v - v_{0}}{t}$
$\therefore v = v_{0} + at$.
$2$. Second Equation of Motion $(x = v_{0}t + \frac{1}{2}at^{2})$:
The displacement $x$ is the area under the $v-t$ graph (trapezium $OABD$).
$x = \text{Area of rectangle } OACD + \text{Area of } \Delta ACB$
$x = (v_{0} \times t) + \frac{1}{2} \times (t) \times (v - v_{0})$
Since $(v - v_{0}) = at$,we get:
$x = v_{0}t + \frac{1}{2}at^{2}$.
$3$. Third Equation of Motion $(v^{2} - v_{0}^{2} = 2ax)$:
The displacement $x$ is the area of the trapezium $OABD$.
$x = \frac{1}{2} \times (\text{sum of parallel sides}) \times \text{height}$
$x = \frac{1}{2} (v_{0} + v) \times t$
Since $t = \frac{v - v_{0}}{a}$,we substitute $t$:
$x = \frac{1}{2} (v + v_{0}) \times \frac{(v - v_{0})}{a}$
$2ax = v^{2} - v_{0}^{2}$
$\therefore v^{2} = v_{0}^{2} + 2ax$.