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

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

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

D) \[a/(v+{{v}_{1}})\] 

Correct Answer: B

Solution :

 Let two boys meet at point C after time 't' from the starting. Then \[AC=vt,\,BC={{v}_{1}}t\] By solving we get \[\sqrt{\frac{{{a}^{2}}}{{{v}^{2}}-v_{1}^{2}}}\]