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]

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                 [CBSE PMT 2005]

TuteeHUB · Accepted Answer

$$a/(v-{{v}_{1}})$$

TuteeHUB · Answer

$$a/\sqrt{{{v}^{2}}+v_{1}^{2}}$$

TuteeHUB · Answer

$$\sqrt{{{a}^{2}}/({{v}^{2}}-v_{1}^{2})}$$

TuteeHUB · Answer

$$a/(v+{{v}_{1}})$$

1.	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.	\[a/(v+{{v}_{1}})\]
Answer» C. \[a/(v-{{v}_{1}})\]

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]

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