Question

Derive the following equation of motion by the graphical method : v² = u² + 2as where the symbols have their usual meanings.

Answer 1

In the given figure, the distance travelled (s) by a body in time (t) is given by the area of the figure OABC which is a trapezium.
Distance travelld = Area of the trapezium OABC
So, Area of trapezium OABC,
$$\begin{gathered} =\frac{(\mathrm{Sum}\mathrm{of}\mathrm{parallel}\mathrm{sides})\left(\mathrm{Height}\right)}{2} \\ =\frac{(\mathrm{OA}+\mathrm{CB})\left(\mathrm{OC}\right)}{2} \end{gathered}$$
Now, (OA + CB) = u + v and (OC) = t.
Putting these values in the above relation, we get:
$s=\left(\frac{u+v}{2}\right)t...\left(1\right)$
Eliminate t from the above equation. This can be done by obtaining the value of t from the first equation of motion.
v = u + at
So,
$t=\frac{v-u}{a}$
Now, put this value of t in equation (1), we get:
$s=\left(\frac{(u+v)(v-u)}{2a}\right)$
On further simplification,
2as = v² – u²
Finally the third equation of motion.
$\boxed{v^2=u^2+2as}$
where
(s) - Displacement
(u) - Initial velocity
(a) - Acceleration
(v) - Final velocity
(t) - Time taken