Q. If the velocity of particle is given by $v = (180 - 16x)^{1/2}\ m/s$, then its acceleration will be
A) Zero
B) $8\ m/s^{2}$
C) $- 8\ m/s^{2}$
D) $4\ m/s^{2}$
Explanation
(c) $v = (180 - 16x)^{1/2}$
As $a = \frac{dv}{dt} = \frac{dv}{dx}.\frac{dx}{dt}$
$\therefore a = \frac{1}{2}(180 - 16x)^{- 1/2} \times ( - 16)\left( \frac{dx}{dt} \right)$
$= - 8(180 - 16x)^{- 1/2} \times v$
$= - 8(180 - 16x)^{- 1/2} \times (180 - 16x)^{1/2} = - 8\ m/s^{2}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.