Question

A particle moves along an arc of a circle of radius $R.$ Its velocity depends on the distance covered $s$ as $v=a\sqrt{s}$, where $a$ is a constant then the angle $\alpha$ between the vector of the total acceleration and the vector of velocity as a function of $s$ will be:

• A. $\tan \alpha =\frac{R}{2s}$
• B. $\tan \alpha =\frac{2s}{R}$
• C. $\tan \alpha =\frac{2R}{s}$
• D. $\tan \alpha =\frac{s}{2R}$

Answer 1

The correct option is B $\tan \alpha =\frac{2s}{R}$

Given:-$v=a\sqrt{s}$

By using the above equation we can find the tangential vector

$w_t=\frac{dv}{dt}=\frac{a^2}{2}$

$w_n=\frac{v^2}{R}=\frac{a^{2}s}{R}$

The relation between w and v is given as,

$\tan \alpha =\frac{w_n}{w_t}=\frac{a^{2}s\times 2}{a^{2}R}$

$\Rightarrow$$\tan \alpha =\frac{2s}{R}$