A particle is moving in a straight line such that its distance `s` at any time `t` is given by `s=(t^4)/4-2t^3+4t^2-7.` Find when its velocity is maximum and acceleration minimum.

Explanation

`S=t^4/4-2t^3+4t^2-7`

`V=(dS)/(dt)=t^3-6t^2+8t`

`a=(dV)/(dt)=(d^2S)/(dt^2)=3t^2-12t+8=0`

`t=(12pmsqrt(144-96))/6`

`t=2pm2/sqrt3`

`(d^2V)/(dt^2)=6t-12`

`t=2+2/sqrt3`

`(d^2V)/(dt^2)=6(2+2/sqrt3)-12`

...