Question

A point moves along an arc of a circle of radius R. Its velocity depends upon the distance covered `s' as $v=A\sqrt{s}$, where a is constant. The angle $\theta$ between the vector of total acceleration and tangential acceleration is:

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

Answer 1

The correct option is D $tan\theta =\frac{2s}{R}$


$v=2\sqrt{s}$ .....(i)
Squaring both sides
$v^2=a^{2}s$ ....(ii)
$a_c=\frac{v^2}{R}=\frac{a^{2}s}{R}$
$a_t=\frac{dv}{dt}$ [multiply and divide RHS by ds]
$a_t=\frac{dv}{ds}\left(\frac{ds}{dt}\right)=\frac{vdv}{ds}$ .....(iii) $[v=\frac{ds}{dt}]$
Differentiate equation(ii)
$2v\frac{dv}{ds}=a^2$
$\frac{vdv}{ds}=\frac{a^2}{2}$ use this result in equation (iii) we get
$a_t=\frac{a^2}{2}$
$tan\theta =\frac{a_c}{a_t}=\frac{a^{2}s\times 2}{R{a}^2}=\frac{2s}{R}$