Question

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

Answer 1

Answer: (B)

Acceleration is the rate of change of velocity and velocity is rate of change of displacement.
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.
i.e., $v=\frac{dx}{dt}$
$=\frac{d}{dt}\left(a_0+\frac{a_{1}t}{2}-\frac{a_2{t}^2}{3}\right)$
$=0+\frac{a_1}{2}-\frac{2a_{2}t}{3}$
$=\frac{a_1}{2}-\frac{2a_{2}t}{3}$
Acceleration = rate of change of velocity
i.e., $a=\frac{dv}{dt}$
$=\frac{d}{dt}(\frac{a_1}{2}-\frac{2a_2}{3}t)$
$=0-\frac{2a_2}{3}=-\frac{2a_2}{3}$