Derive the kinematic equations of motion for constant acceleration.

Explanation

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 velocity of the body at a later time t.
Velocity - time relation:
(i) 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 = a int_(0)^(t) dt rArr [v]_u^v = a [t]_0^t` .....(1)
Displacement - time relation
(ii) 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 + 1/2 at^2` ...(2)
Velocity - displacement relation (iii) 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
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)`
` therefore v^2 = u^2 + 2as` ...(3)
We can also derive the displacement s in terms of initial velocity u and final velocity v. From equation we can write,
at = v- u
Substitute this in equation, we get
` s = ut + 1/2 (v - u) t`
`s = ((u+v)t)/(2) `