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

Given:

The distance covered by a particle at time ' `t` ' is given by,


`S=frac{t^{4}}{4}-2 t^{3}+4 t^{2}-7`


We know that velocity of a particle is given by `frac{d S}{d t}` and acceleration of a particle is given by `frac{d^{2} S}{d t^{2}}`.


`Rightarrow v=frac{d S}{d t}=frac{d(frac{t^{4}}{4}-2 t^{3}+4 t^{2}-7)}{d t} `

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