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 \[{{s}_{\min }}=BC=AB\cos \theta =\frac{a{{v}_{2}}}{v}\]You need to login to perform this action.
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