Question

Derive first equation of motion by graphical method.

Answer 1

Let us consider an object moving in a straight line with a constant acceleration. For such a situation the first equation of motion gives the final velocity (v) after time t given that the object is having an initial velocity of u and constant acceleration of a. Now for deriving it, let us consider the following velocity-time graph:
The initial velocity of the body, $u=OA.................\left(1\right)$
The final velocity of the body, $v=BC....................\left(2\right)$
From the graph $BC=BD+DC$
Therefore, $v=BD+DC..............................\left(3\right)$
Again $DC=OA$
So, $v=BD+OA$
Now from equation (1), $OA=u$
So, $v=BD+u......................................\left(4\right)$
We should find the value of BD now
Now, from the velocity-time graph, if we calculate the change in velocity to change in time, then we can find the acceleration, a. This is nothing but the slope of the graph.
Thus,
Acceleration, a = Slope of line AB
$a=\frac{BD}{AD}$
But $AC=OC=t$
Hence we get
$a=\frac{BD}{t}$
$BD=at$
Now putting this in equation (4) we get
$v=u+at$