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:            [AIPMT (S) 2005]

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

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