(a) If the initial velocity of a particle is u and collinear acceleration at any time `t` is at, calculate the velocity of the particle after time `t`.
(b) A particle moves along a straight line such that its displacement at any time t is given by `s = (t^(3) - 6 t^(2) + 3t + 4)m`. What is the velocity of the particle when its acceleration is zero?

Explanation

By definition acceleration `= (dv//dt)`
So, `(dv)/(dt) = at` (given)
or `int_(u)^(v) dv = int_(0)^(t) at`
or `v - u = (1)/(2) a t^(2)`
or `v = u + (1)/(2) a t^(2)`
(b) As according to given problem,
`s = t^(3) - 6t^(2) + 3t + 4`
instantaneous velocity
`u (ds)/(dt) = 3t^(2) - 12 t + 3`
and acceleration
`a = (dv)/(dt) = (d^(2)s)/(dt^(2)) = 6 t - 12`....(i)
So, acceleration will be zero when `6t - 12 = 0`, i.e., `t = 2` sec
And so the velocity when acceleration is zero, i.e., at `t = 2` sec from Eqn. (i), will be
`v = 3 xx 2^(2) - 12 xx 2 + 3 = - 9m//s`
[Negative velocity means that body is moving towards the origin, i.e., as time increases displacement decreases.]