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

To solve the problem, we need to find when the velocity of a particle is maximum and when its acceleration is minimum based on the given distance function \( s(t) = \frac{t^4}{4} - 2t^3 + 4t^2 - 7 \).

### Step 1: Find the Velocity Function
The velocity \( v(t) \) is the first derivative of the distance function \( s(t) \) with respect to time \( t \).

\[ v(t) = \frac{ds}{dt} = \frac{d}{dt}\left(\frac{t^4}{4} - 2t^3 + 4t^2 - 7\right) \]
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