question_answer

Two boys are standing at the ends A and B of a ground, where AB = a. The boy at B starts running in a direction perpendicular to AB with velocity \[{{v}_{1}}\]. The boy at A starts running simultaneously with velocity v and catches the other boy in a time \[t,\] where t is

A) \[\frac{a}{\sqrt{{{v}^{2}}+v_{1}^{2}}}\] 

B)        \[\sqrt{\frac{{{a}^{2}}}{{{v}^{2}}-v_{1}^{2}}}\]   

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

D)        \[\frac{a}{(v+{{v}_{1}})}\]

Correct Answer: B

Solution :

Distance covered by boy A in time \[t\] \[AC=vt\]                                            ?(i) Distance covered by boy B in time t \[BC={{v}_{1}}t\]                                              ?(ii) Using Pythagorus theorem \[A{{C}^{2}}=A{{B}^{2}}+B{{C}^{2}}\] or            \[{{(vt)}^{2}}={{a}^{2}}+{{({{v}_{1}}t)}^{2}}\] or            \[{{v}^{2}}{{t}^{2}}-v_{1}^{2}{{t}^{2}}={{a}^{2}}\] or            \[{{t}^{2}}({{v}^{2}}-v_{1}^{2})={{a}^{2}}\] \[\therefore \]  \[t=\sqrt{\frac{{{a}^{2}}}{({{v}^{2}}-v_{1}^{2})}}\]