Question

A particle moves along a straight line and its velocity at a distance 'x' from the origin is $k\sqrt{a^2-x^2}$. Then acceleration of the particle is

• A. $k$
• B. $-k^2$
• C. $kx$
• D. $-k^{2}x$

Answer 1

The correct option is D $-k^{2}x$

Given,

Velocity,

$v=k\sqrt{a^2-x^2}$

$\frac{dv}{dx}=-\frac{kx}{\sqrt{a^2-x^2}}$

Acceleration,

$a=v\times \frac{dv}{dx}$

$=k\sqrt{a^2-x^2}\times (-\frac{kx}{\sqrt{a^2-x^2}})$

$\therefore a=-k^{2}x$