question_answer

The distance between two moving particles at any time is a. If v be their relative velocity, and \[{{v}_{1}}\] and \[{{v}_{2}}\] be the components of v along and perpendicular to a, the minimum distance between them is

A) \[\frac{av}{2}\]            

B)        \[\frac{av}{{{v}_{1}}}\]                     

C) \[\frac{a{{v}_{1}}}{v}\]                    

D)       \[\frac{a{{v}_{2}}}{v}\]

Correct Answer: D

Solution :

[d] \[{{v}^{2}}=v_{1}^{2}+v_{2}^{2}\] \[\tan \theta =\frac{{{v}_{1}}}{{{v}_{2}}}\] \[\therefore \,\,\,\,\,\,\,\,\,\,\,\,\,\cos \theta =\frac{{{v}_{2}}}{\sqrt{v_{1}^{2}+v_{2}^{2}}}=\frac{{{v}_{2}}}{v}\] and \[\sin \theta =\frac{{{v}_{1}}}{\sqrt{v_{1}^{2}+v_{2}^{2}}}=\frac{{{v}_{1}}}{v}\] Minimum distance between A and B is \[{{s}_{\min }}=BC=AB\cos \theta =\frac{a{{v}_{2}}}{v}\]