Derive the equations of uniformaly accelerated motion by graphical method.

Explanation

Now at `t=0` the velocity of the particle be `v_(0)` and at t =t the velocity of the particle be v.

`therefore` Acceleration a = slop of line AB
`= (v-v _(0))/( t -0)`
`therefore a = (v - v _(0))/(t )`
`therefore v = v _(0) + at " "...(1)`
Now displacement of a particle during time t is equal to the area of region OABCD in `v to t` graph.
`therefore x =` Area of rectangle OACD + Area of `Delta ACB`
`=v_(0) t + 1/2 (v - v _(0)) t`
`therefore x = v _(0) t + 1/2 at ^(2) " "...(2)`
but average velocity,
` x = ((v + v _(0))/( 2 )) t " "...(3)`
For, equation (3) and equatin (1),
`t = (v-v_(0))/(a)`
`therefore x = ((v + v _(0))/(z )) ((v - v _(0))/(a ))`
`therefore v ^(2) -v _(0) ^(2) = 2ax`
Above equation can be written as follows :
`v = v _(0) + at`
`x =x _(0) +v_(0) t + 1/2 at ^(2)`
`2a (x -x_(0)) =v^(2) -v_(0) ^(2)`