Derive the kinematic equations of motion for constant acceleration.

#### Solution

Consider an object moving in a straight line with uniform or constant acceleration ‘a’. Let u be the velocity of the object at time t = 0, and v be the velocity of the body at a later time t.

**Velocity – time relation:**

(1) The acceleration of the body at any instant is given by the first derivative of the velocity with respect to time,

a = `"dv"/"dt"` or dv = a dt

Integrating both sides with the condition that as time changes from 0 to t, the velocity changes from u to v. For the constant acceleration,

`int_u^v dv = int_0^t a dt = aint_0^t dt ⇒ [v]_u^v =a[t]_0^t`

v - u = at or v = u + at .............(i)

(2) The velocity of the body is given by the first derivative of the displacement with respect to time.

v = `"ds"/"dt"` or ds = vdt

and since v = u + at,

we get ds = (u+ at ) dt

Assume that initially at time t = 0, the particle started from the origin. At a later time t, the particle displacement is s. Further assuming that acceleration is time-independent, we have

`int_0^s ds = int_0^t u dt + int_0^t at dt "or" s = ut + 1/2at^2` .................(ii)

**Velocity – displacement relation:**

(3) The acceleration is given by the first derivative of velocity with respect to time.

a = `"dv"/"dt" = "dv"/"ds" = "ds"/"dt" = "dv"/"ds"`V [since ds/dt = v] where s is displacement traverse

This is rewritten as a = `1/2 "dv"^2/"ds"` or ds = `1/(2a) d(v^2)` Integrating the above equation, using the fact when the velocity changes from u^{2} to v^{2}, displacement changes from 0 to s, we get

`int_0^s ds = int_u^v 1/(2a) d(v^2)`

s = `1/(2a)(v^2 - u^2)`

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

We can also derive the displacement 5 in terms of initial velocity u and final velocity

v. From the equation we can

write,

at = v – u

Substitute this in the equation, we get

`s = ut + 1/2(v - u)t`

s = `((u + v)t)/2`