A particle moving in a straight line is said to be uniformly accelerated, if it makes equal changes in velocities in equal intervals of time, however small the time interval may be.

In uniformly accelerated motion, instantaneous acceleration is constant and is independent of time.

`text(Derivation of equations for uniformly accelerated motion)`

`text((a))` `a= (dv)/(dt)`

`int_(v=0)^(v=v) dv =a int_(t=0)^(t=t)dt [v]_u^v = a[t]_0^t`

`v = u +at .................(1)`

`text((b))` `v= (ds)/(dt)`

From Eq. (1) `v = u + a t`

`(ds)/(dt) = u +at`

`int_(s=0)^(s=s) ds = int_(t=0)^(t=t)(u+at) dt`

`int_(s=0)^(s=s)ds =int_(t=0)^(t=t) u dt +int_(t=0)^(t=t)at dt`

`s= ut +1/2at^2...............(2)`


`text((c))` Let at `t=0 , x=x_0`

`t=t, x=x`

`v=(dx)/(dt)`

`(dx)/(dt) = u +at`

`int_(x=x_0)^(x=x) dx =int_(t=0)^(t=t)(u+at)dt`

`x=x_0 +ut +1/2 at^2`

or `x-x_0 = ut +1/2 at^2 ............(3)`

(Here `s = x- x_0 =` displacement or change in position)


`text((d))` `a= v((dv)/(ds))`

`int_(v=u)^(v=v) vdv =a int_(s=0)^(s=s)ds `

`v^2 =u^2 +2as............(4)`

`text((e))` `a= v((dv)/(dx))`

`int_(v=u)^(v=v) vdv = a int_(x=x_0)^(x=x)dx `

`v^2 = u^2 +2a(x-x_0)...........(5)`


`text((f))` Eliminating t between Eq. (1) and Eq. (2), we get

`s= (u+v)/2 xx t............(6)`


`text((g))` Displacement in `n^(th)` second for a uniformly accelerated body.

Let total displacement of body in n seconds be `S_n`

`S_n = un +1/2 an^2`

Let total displacement of body in `(n - 1)` seconds be `S_(n-1)`

`S_(n-1) = u(n-1) +1/2 a (n-1)^2`

Then displacement of body in `n^(th)` second will be

`S_(n^(th)) = S_n - S_(n-1)`

`S_(n^(th)) = un + 1/2 an^2 -[u(n-1) +1/2a(n-1)^2]`

`S_(n^(th)) = u + a[n- 1/2]....................(7)`