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 y 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            \[{{t}^{2}}({{v}^{2}}-v_{1}^{2})={{a}^{2}}\]                         \[\therefore t=\sqrt{\frac{{{a}^{2}}}{({{v}^{2}}-v_{1}^{2})}}\]