Derive the kinematic equations of motion for constant acceleration.

Explanation

Let us 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
Let v be velocity of the body at a later time t.
Velocity - time relation : The acceleration of the body at any instant is given by
` = (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,
`underset(u)overset(v)intdv = underset(0)overset(t)inta dt = a underset(0)overset(t)int dt rArr [v]_(u)^(v)a [t]_(0)^(t)`
`v - u = at (or) v = u + at " ". . . (1)`
Displacement-time relation : The velocity of the body is given by
`v = (ds)/(dt) or ds = vdt`
But , v = u + at,
`:. ` ds = (u + at) dt
Let us 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 get
`underset(0)overset(z)int ds = underset(0) overset(t)int u dt = underset(0)overset(t) int at dt (or)`
`:. s = ut +(1)/(2) at^(2)` . . . (2)
Velocity- displacement . relation : The acceleration is given by the first derivative of velocity with respect to time .
`a = (dv)/(dt) = (dv)/(ds)(ds)/(dt) = (dv)/(ds) v`
Let is displacement traversed [since ds/dt = v]
`:. a = (1)/(2) (dv^(2))/(ds)`
or ds ` = (1)/(2a) d (v^(2))`
When the velocity changes from `u^(2) "to"v^(2)` , displacement changes from 0 to s, integrating the above equation, we get
`underset(0)overset(s)int ds = underset(m) overset(v) int (1)/(2a) d(v^(2))`
`:. s = (1)/(2a) (v^(2) - u^(2))`
`rArr :. v^(2) = u^(2) + 2as " ". . . (3)`
Let us derive the displacement s in terms of initial velocity u and finla velocity v .
From the equation (1) we can write,
at = v - u
Substitute this in equation (2), we get
`s = ut (1)/(2) (v -u) t`
`:. s = ((u +v)t)/(2)`