question_answer

A point moves along an arc of a circle of radius R. Its velocity depends upon the distance covered V as v=aVs, where a is constant. The angle e 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) \[\tan \,\theta =\frac{2s}{2R}\]

Correct Answer: D

Solution :

\[\operatorname{v}=a\sqrt{s}\]               ....(1) squaring both sides \[{{\operatorname{v}}^{2}}={{a}^{2}}s\]                              ....(2) \[{{\operatorname{a}}_{c}}=\frac{{{v}^{2}}}{R}=\frac{{{a}^{2}}s}{R}\] \[{{\operatorname{a}}_{t}}=\frac{dv}{dt}\]       [multiply and divide RHS by ds] \[{{\operatorname{a}}_{t}}=\frac{dv}{dt}\,\,\left( \frac{ds}{dt} \right)\frac{vdv}{ds}\]      ...(3)\[\left[ d=\frac{ds}{dt} \right]\] Differentiate equation (2) \[2v.\frac{dv}{ds}={{a}^{2}}\] \[\frac{vdv}{ds}=\frac{{{a}^{2}}}{2}\] use this result in equation (3) we get \[{{\operatorname{a}}_{t}}=\frac{{{a}^{2}}}{a}\] \[\tan \,\theta \,=\frac{{{a}_{c}}}{{{a}_{t}}}=\frac{{{a}^{2}}s\times 2}{R{{a}^{2}}}=\frac{2s}{R}\]