Question

Derive an equation for position-velocity relation $(2as=v^2-u^2)$ by graphical method

Answer 1

Let the initial velocity of the object = u

Let the object is moving with uniform acceleration, a.

Let object reaches at point B after time, t and its final velocity becomes, v

Draw a line parallel to x-axis DA from point, D from where object starts moving.

Draw another line BA from point B parallel to y-axis which meets at E at y-axis.

The distance covered by the object moving with uniform acceleration is given by the area of trapezium ABDO

Therefore,

Area of trapezium ABDOE $=\frac{1}{2}\times$ (sum of parallel sides + distance between parallel sides)

$Distance\left(S\right)=\frac{1}{2}(DO+BE)\times OE$

$S=\frac{1}{2}(u+v)\times t...............\left(i\right)$

we know that,

$a=\frac{v-u}{t}$

from above equation we can say,

$t=\frac{v-u}{a}...........\left(ii\right)$

After substituting the value of $t$ from equation$\left(ii\right)$ in equation $\left(i\right)$

$S=\frac{1}{2a}(u+v)(v-u)$

$2aS=(u+v)(v-u)$

$2aS=v^2-u^2$

Hence Proved.