Q. The position $x\ $of a particle varies with time $t\ $as $x = at^{2} - bt^{3}.$ The acceleration of the particle will be zero at time $t\ $equal to
A) $\frac{a}{b}$
B) $\frac{2a}{3b}$
C) $\frac{a}{3b}$
D) $Zero$
Explanation
(c) $\frac{dx}{dt} = 2at - 3bt^{2} \Rightarrow \frac{d^{2}x}{dt^{2}} = 2a - 6bt = 0 \Rightarrow t = \frac{a}{3b}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.