Question

A particle moves in a straight line according to the law $v^2=4a(x\sin x+\cos x)$ where $v$ is the velocity of a particle at a distance $x$ from the fixed point. Then the acceleration is

• A. $ax\cos x$
• B. $2ax\cos x$
• C. $ax\sin x$
• D. $2ax\sin x$

Answer 1

The correct option is B $2ax\cos x$

Given: $v^2=4a(x\sin x+\cos x)$
We know, Acceleration $=\frac{dv}{dt}$
$\Rightarrow 2v\frac{dv}{dt}=4a(x\cos x\frac{dx}{dt}+\sin x\frac{dx}{dt}-\sin x\frac{dx}{dt})$
$\Rightarrow \frac{dv}{dt}=2ax\cos x(\because \frac{dx}{dt}=v)$