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Standard XII

Physics

Velocity

Two boys are ...

Question

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

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Solution

The correct option is B
Let two boys meet at point C after time 't' from the starting Then AC=vt. BC= $v_{1} t$

$(A C)^{2} = (A B)^{2} + (B C)^{2} \Rightarrow v^{2} t^{2} = a^{2} + v_{1}^{2} t^{2}$

By solving we get t= $\sqrt{\frac{a^{2}}{v^{2} - v_{1}^{2}}}$

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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 v1). The boy at A starts running simultaneously with velocity v and catches the other boy in a time t, where t is

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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

The correct option is B
Let two boys meet at point C after time 't' from the starting Then AC=vt. BC= $v_{1} t$

$(A C)^{2} = (A B)^{2} + (B C)^{2} \Rightarrow v^{2} t^{2} = a^{2} + v_{1}^{2} t^{2}$

By solving we get t= $\sqrt{\frac{a^{2}}{v^{2} - v_{1}^{2}}}$