Question

A particle moves in a circular path such that its speed V varies with distance as V is equal to Alpha Root over S. Where Alpha is a positive constant. find the acceleration of the particle after travelling a distance S ?​

Answer 1

Answer:

$\frac{\alpha^2}{2}$

Explanation:

1) We have,

Velocity of a particle moving in a circular path,with its speed varying with distance as follows:

$v=\alpha \sqrt{s}$

Then, we know acceleration of that particle is given by:

$a=\frac{dv}{dt}\\ \\=\frac{dv}{ds}*\frac{ds}{dt}=\frac{dv}{ds}v$

That is,

$v=\alpha \sqrt{s}\\ \\ =>\frac{dv}{ds}=\alpha*\frac{1}{2}\frac{1}{\sqrt{s}}=\frac{\alpha}{2\sqrt{s}}$

2) And Finally, the acceleration is given by :

$a=\alpha \sqrt{s}*\frac{1}{2\sqrt{s}}\alpha =\frac{\alpha^2}{2}$

Hence, the acceleration of a particle after travelling a distance S, is

$\boxed{\frac{\alpha^2}{2}}$