Derive the relation `v ^(2) - u ^(2) = 2a S,`

Explanation

We know tht, are aunder v - t graph gives the distance travelled by the body. Form distance travelled = area of trapezium OPQS
`implies S = (1)/(2) (O P + QS) xx PR .....(i)`
Also acceleration a = slope of v - t graph `implies a = ( QR)/( PR) = ( QS - RS)/( PR)`
or `PR = ( v - u )/(a) `
Here ` OP = u and QS = v`
Substituting in (i), we get
or `S = (1)/(2) (v + u ) ((v - u )/( a )) = (( v ^(2) - u ^(2))/( 2 a )) or v ^(2) - u ^(2) =2aS`