A partical is moving in a straight line such that `s=t^(3)-3t^(2)+2`, where `s` is the displacement in meters and `t` is in seconds. Find the
(a) velocity at `t=2s`, (b) acceleration at `t=3s`,
( c) velocity when acceleration is zero and
(d) acceleration when velocity is zero.

Explanation

`s=t^(3)-3t^(2)+3t+2`
`v=(ds)/(dt)=3t^(2)-6t+3`
`a=(dv)/(dt)=6t-6`
(a) At `t=2 s, v=3(2)^(2)-6(2)+3=3 m//s `
(b)At` t=1 s, a=6(3)-6=12 m//s^(2)`
(c ) Acceleration `a=0implies6t-6=0impliest=1 s `
`At t=1 s, v=3(1)^(2)-6(1)+3=0`
(d) When velocity is zero, `v=3t^(2)-6t+3=3(t^(2)=2t+1)=0`
`(t-1)^(2)=0impliest=1 s `
At `t=1 s, a=6(1)-6=0`