Derive the equation of motion
`x=v_0t+1/2at^2` using appropriate graph.

Explanation

The velocity -time graph of a body moving with initial velocity `u` and with uniform acceleration `a` as shown. Let `v` Be the velocity of the body after a time t.
In v-t graph area of velocity -time graph = total displacement travelled by it. Area under velocity - time graph = area of OABCD
`therefore` Area of Rectangular part OACD
=Area of OACD + Area of ABC.
`A_1=OAtimesOD=v_0.t............(1)`

2) Area of triangle ABC=`A_2`
`A_2=1/2Base times height`
`=1/2ACtimesBC`
`=1/2t(v-v_0)`
But `v-v_0`= at
`thereforeA_2=1/2t.at=1/2at^2`.
`therefore` Total area under graph `=s=A_1+A_2`
`s(n)=v_0t+1/2at^2`
`therefores=ut+1/2at^2` is graphically proved.