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                 [CBSE PMT 2005]

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 \[AC=v\,t\],  \[BC={{v}_{1}}t\]             \[{{(AC)}^{2}}={{(AB)}^{2}}+{{(BC)}^{2}}\] Þ  \[{{v}^{2}}{{t}^{2}}={{a}^{2}}+v_{1}^{2}{{t}^{2}}\]             By solving we get \[t=\sqrt{\frac{{{a}^{2}}}{{{v}^{2}}-v_{1}^{2}}}\]