Q. A particle moves along a straight line such that its displacement at any time $t$ is given by $s = t^{3} - 6t^{2} + 3t + 4.$ The velocity when its acceleration is zero is
A) $2\ {ms}^{- 1}$
B) $12\ {ms}^{- 1}$
C) $- 9\ {ms}^{- 1}$
D) $2\ {ms}^{- 1}$
Explanation
(c) Given, $\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ s = t^{3} - 6t^{2} + 3t + 4$
Velocity $\ \ \ \ \ \ \ \ \ v = \frac{ds}{dt} = 3t^{2} - 12t + 3\ \ \ \ \ \ \ \ \ \ \ldots(i)$
Acceleration $a = \frac{dv}{dt} = 6t - 12\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ldots(ii)$
Since, acceleration is zero, so, $6t - 12 = 0,\ or\ t = 2\ s$
So, velocity $v\ at$
$\ t = 2\ s,\ is = 3 \times 2^{2} - 12 \times 2 + 3 = - 9\ {ms}^{- 1}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.