. Displacement ' $s$ ' of a particle, in meters, at any time ' $t$ ' $s$ in expressed as $s=\frac{t^3}{3}-6…
. Displacement ' $s$ ' of a particle, in meters, at any time ' $t$ ' $s$ in expressed as $s=\frac{t^3}{3}-6 t$. Find the acceleration at a time when the velocity vanishes,
$6 \mathrm{~m} \cdot \mathrm{s}^{-2}$
$2 \sqrt{6} \mathrm{~m} \cdot \mathrm{s}^{-2}$
$12 \mathrm{~m} \cdot \mathrm{s}^{-2}$
$6 \sqrt{6} \mathrm{~m} \cdot \mathrm{s}^{-2}$
Solution
Given, displacement of a particle at time $t$ as
$
\begin{gathered}
s=\frac{t^3}{3}-6 t \\
\therefore \text { Velocity }=\frac{d s}{d t}=t^2-6=v
\end{gathered}
$
and as velocity vanishes, so $v=0 \Rightarrow t=\sqrt{6} \mathrm{sec}$
Now, the acceleration $a=\frac{d v}{d t}=2 t$
$\therefore$ Acceleration $a$ at $t=\sqrt{6} \mathrm{sec}$ is $2 \sqrt{6} \mathrm{~m} / \mathrm{sec}^2$