Q. The displacement of particle is given by$x = a_{0} + \frac{a_{1}t}{2} - \frac{a_{2}t^{2}}{3}$ What is its acceleration?

A) $\frac{2a_{2}}{3}$
B) $- \frac{2a_{2}}{3}$
C) $a_{2}$
D) Zero
Explanation

(b) The displacement equation is given by
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ x = a_{0} + \frac{a_{1}t}{2} - \frac{a_{2}t^{2}}{3}$
Velocity = rate of change of displacement
$ie,\ \ v = \frac{dx}{dt}$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = \frac{d}{dt}\left( a_{0} + \frac{a_{1}t}{2} - \frac{a_{2}t^{2}}{3} \right)$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = 0 + \frac{a_{1}}{2} - \frac{2a_{2}t}{3}$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = \frac{a_{1}}{2} - \frac{2a_{2}t}{3}$
Acceleration = rate of change of velocity
$ie,\ \ a = \frac{dv}{dt}$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = \frac{d}{dt}\left( \frac{a_{1}}{2} - \frac{2a_{2}}{3}\ t \right)$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = 0 - \frac{2a_{2}}{3}$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = - \frac{2a_{2}}{3}$

Dr. Abubacker
Reviewed by Dr. Abubacker

Physics Educator & NEET/JEE Specialist

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