A particle moves along a staight line such that its displacement at any time t is given by `s=t^3-6t^2+3t+4m`. Find the velocity when the acceleration is 0.

Explanation

` S=t^3 - 6 t^2+ 3 t + 4,` Differentiating it w.r.t. to (t) we have ltbRgt (dS) /(dt) = velocity ` =v= 3 t^2 - 12 t + 3` …(i)
Differntiatng it angin w.e.t. we have
`(dv)/(dt)= acceleration =a6 t-12`
Accelration, ` a=0 ` if ` 6t - 12 =0`
or ` t=2 seconds`
Putting this value in (i) , we get ltbRgt ` v= 3 xx 2^2- 12- 12 xx 2 + 3=- 9 m//a`.