Show that `v^(2)=v_(0)^(2)+2ax` graphical method.

Explanation

Let `'v_(0)'` be the initial velocity of a particle describing uniform accelerated motion 'a'.
Slope of the line `AB=a=tan theta=(BC)/(AC)` where `BC=v-v_(0) and AC=t`
`"i.e. "a=(v-v_(0))/(t)" ….(1)"`
But area of trapezium `=x=1//2 xxODxx(OA+BD)`
`"i.e. "x=1//2t(v_(0)+v)" ...(2)"`
From (1),`" "t=(v-v_(0))/(a)`
Hence (2) may be written as `2s=((v-v_(0))/(a))(v+v_(0))`
`"i.e. "v^(2)-v_(0)^(2)=2ax`
`therefore" "v^(2)=v_(0)^(2)+2ax`

Note : For vertical motion,
`a=-g` and `x=-y` for downward motion and `a=-g, x=+y` for upward motion.