If $s=t^{3}-4 t^{2}+5$ describes the motion of a particle, then its velocity when the acceleration vanishes, is
(a) $\frac{16}{9}$ unit / sec
(b) $-\frac{32}{3}$ unit/sec
(c) $\frac{4}{3}$ unit/sec
(d) $-\frac{16}{3}$ unit/sec
(d) $-\frac{16}{3}$ unit / sec
According to the question,
$s=t^{3}-4 t^{2}+5$
$\Rightarrow \frac{d s}{d t}=3 t^{2}-8 t$
$\Rightarrow \frac{d^{2} s}{d t^{2}}=6 t-8$
$\Rightarrow 6 t-8=0$ $\left[\right.$ As velocity deminishes, then $\left.\frac{d^{2} s}{d t^{2}}=0\right]$
$\Rightarrow t=\frac{4}{3}$
Now, $\left(\frac{d s}{d t}\right)_{t=\frac{4}{3}}=3\left(\frac{4}{3}\right)^{2}-8\left(\frac{4}{3}\right)$
$\Rightarrow \frac{d s}{d t}=\frac{16}{3}-\frac{32}{3}$
$\Rightarrow \frac{d s}{d t}=-\frac{16}{3}$ unit $/ \mathrm{sec}$
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