#### Question

A particle is moving in a straight line such that its distance at any time *t* is given by S = \[\frac{t^4}{4} - 2 t^3 + 4 t^2 - 7 .\] Find when its velocity is maximum and acceleration minimum.

#### Solution

\[\text { Given }: \hspace{0.167em} s = \frac{t^4}{4} - 2 t^3 + 4 t^2 - 7\]

\[ \Rightarrow v = \frac{ds}{dt} = t^3 - 6 t^2 + 8t\]

\[ \Rightarrow a = \frac{dv}{dt} = 3 t^2 - 12t + 8\]

\[\text { For maximum or minimum values of v, we must have }\]

\[\frac{dv}{dt} = 0\]

\[ \Rightarrow 3 t^2 - 12t + 8 = 0\]

\[\text { On solving the equation, we get }\]

\[t = 2 \pm \frac{2}{\sqrt{3}}\]

\[\text { Now }, \]

\[\frac{d^2 v}{d t^2} = 6t - 12\]

\[\text {At t } = 2 - \frac{2}{\sqrt{3}}: \]

\[\frac{d^2 v}{d t^2} = 6\left( 2 - \frac{2}{\sqrt{3}} \right) - 12\]

\[ \Rightarrow \frac{- 12}{\sqrt{3}} < 0\]

\[\text { So, velocity is maximum at t } = \left( 2 - \frac{2}{\sqrt{3}} \right) . \]

\[\text { Again }, \]

\[\frac{da}{dt} = 6t - 12\]

\[\text { For maximum or minimum values of a, we must have }\]

\[\frac{da}{dt} = 0\]

\[ \Rightarrow 6t - 12 = 0\]

\[ \Rightarrow t = 2\]

\[\text { Now,} \]

\[\frac{d^2 a}{d t^2} = 6 > 0\]

\[\text { So, acceleration is minimum at t }=2.\]