Question

A particle moving along the $x-$ axis has an instantaneous speed given by $v^2=108-9x^2$. 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 $-9x{\text{m/s}}^2$

We have, $v^2=108-9x^2\Rightarrow v=\sqrt{108-9x^2}$ and we know, $\frac{dx}{dt}=v$

$a=\frac{dv}{dt}=\frac{dv}{dx}\cdot \frac{dx}{dt}$ [Chain rule]
$a=\frac{dv}{dx}\cdot v\Rightarrow a=v\frac{dv}{dx}$
So, $a=v\cdot \frac{d}{dx}\left(\sqrt{108-9x^2}\right)$
$=v\cdot \frac{-18x}{2\sqrt{108-9x^2}}$
$=\sqrt{108-9x^2}\cdot \frac{-9x}{\sqrt{108-9x^2}}$
$\therefore a=-9x{\text{m/s}}^2$
Hence, option A is correct.