Question

A particle is moving along x-axis whose instantaneous speed is $v=\sqrt{108-9x^2}$. The acceleration of the particle is

• A. $-9x{\text{m/s}}^2$
• B. $-18x{\text{m/s}}^2$
• C. $\frac{-9x}{2}{\text{m/s}}^2$
• D. None of these

Answer 1

The correct option is A $-9x{\text{m/s}}^2$

Given,
$v=\sqrt{108-9x^2}$
$a=\frac{dv}{dt}$
$a=\frac{dv}{dx}.\frac{dx}{dt}$
$=\frac{d\left(\sqrt{108-9x^2}\right)}{dx}.\frac{dx}{dt}$
$a=\frac{(-18x)}{2\sqrt{108-9x^2}}.\sqrt{108-9x^2}=-9x{\text{m/s}}^2$