Q. If the distance ‘s’ metres traversed by a particle in $t$ seconds is given by $s = t^{3} - 3t^{2}$, then the velocity of the particle when the acceleration is zero, in metre/second is

A) 3
B) $- 2$
C) $- 3$
D) 2
Explanation

(c)
Given , $s = t^{3} - 3t^{2}$
$\Rightarrow \ \frac{ds}{dt} = 3t^{2} - 6t$ …(i)
$\Rightarrow \ \frac{d^{2}}{dt^{2}} = 6t - 6 = 0$ [given]
$\Rightarrow \ \ t = 1$
On putting value of $t = 1\ in\ Eq.(i),we\ get$
$\frac{ds}{dt} = 3 \times 1 - 6 \times 1 = - 3m/s$

Dr. Abubacker
Reviewed by Dr. Abubacker

Physics Educator & NEET/JEE Specialist

Ph.D. in Physics with 19+ years of teaching experience.