question_answer

Three particles start moving simultaneously from a point on a horizontal smooth plane. First particle moves with speed \[{{V}_{1}}\] towards east, second particle moves towards north with speed \[{{V}_{2}}\] and third one moves towards north east. The velocity of the third particle, so that the three alway lie on a straight line, is:

A)  \[\frac{{{V}_{1}}+{{V}_{2}}}{2}\]                    

B)  \[\sqrt{{{V}_{1}}{{V}_{2}}}\]

C)  \[\frac{{{V}_{1}}{{V}_{2}}}{{{V}_{1}}+{{V}_{2}}}\]         

D)  \[\sqrt{2}\frac{{{V}_{1}}{{V}_{2}}}{{{V}_{1}}+{{V}_{2}}}\]

Correct Answer: D

Solution :

Using the formula, slope of a line passing through \[({{x}_{1}},{{y}_{1}})\,({{x}_{2}},{{y}_{2}})\] and \[({{x}_{3}},{{y}_{3}})\]is \[\frac{{{Y}_{2}}-{{Y}_{1}}}{{{X}_{2}}-{{X}_{1}}}=\frac{{{Y}_{1}}-{{Y}_{3}}}{{{X}_{1}}-{{X}_{3}}}\] We have   \[\frac{{{V}_{2}}+-O}{V-{{V}_{1}}t}=\frac{0-({{\upsilon }_{3}}|\sqrt{2})t}{{{V}_{1}}t-({{V}_{3}}/\sqrt{2})t}\] \[\Rightarrow \]               \[{{V}_{3}}=\frac{\sqrt{2}{{V}_{1}}{{V}_{2}}}{{{V}_{1}}+{{V}_{2}}}\]