Q. A particle moves along a straight line according to the law$\ s$ $= 16 - 2t + 3t^{3}$ , where $s\$a metre is the distance of the particle from a fixed point at the end of $t$ seconds. The acceleration of the particle at the end of 2 s is
A) $36m/s^{2}$
B) $34m/s^{2}$
C) 36m
D) None of these
Explanation
(a)
Given , $\ s = 16 - 2t + 3t^{3}$
$\Rightarrow \ \ \ \ \ \ \ \ \ \ \ \ \frac{ds}{dt} = - 2 + 9t^{2}\ \ \ \Rightarrow \frac{d^{2}s}{dt^{2}} = 18t$
At $t = 2s,\$acceleration a$= \frac{d^{2}s}{dt^{2}} = 18 \times 2 = 36m/s^{2}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.