Question

Derive the kinematic equations of motion for constant acceleration.

Answer 1

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 u² to v², 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`