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

Correct answer: D. (a^2)/(v^2-v_1^2) If boy A catches boy B in time t , aligned & (vt)^2=(v_1 t)^2+a^2 & t^2= a^2 v^2-v_1^2 & t= a v^2-v_1^2 aligned )

NEET 2005

Exam: NEET
Year: 2005
Subject: Physics
Chapter: Motion in Two Dimensions

NEET 2005 · Physics · Motion in Two Dimensions

Two boys are standing at the ends A and B of a ground where AB=a ....

Two boys are standing at the ends

\mathrm{A}

and

\mathrm{B}

of a ground where

\mathrm{AB}=a

. The boy at

\mathrm{B}

starts running in a direction perpendicular to

\mathrm{AB}

with velocity

v_1

. The boy at A starts running simultaneously with velocity

v

and catches the other in time

t

, where

t

is:

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

Options

A $\frac{a}{\sqrt{v^2+v_1^2}}$
B $\frac{a}{v+v_1}$
C $\frac{a}{v-v_1}$
D $\sqrt{\frac{a^2}{v^2-v_1^2}}$

Correct answer

D. $\sqrt{\frac{a^2}{v^2-v_1^2}}$

Step-by-step solution

If boy

A

catches boy

B

in time

t

\begin{aligned} & \Rightarrow(\mathrm{vt})^2=\left(\mathrm{v}_1 \mathrm{t}\right)^2+\mathrm{a}^2 \\ & \Rightarrow \mathrm{t}^2=\frac{\mathrm{a}^2}{\mathrm{v}^2-\mathrm{v}_1^2} \\ & \Rightarrow \mathrm{t}=\frac{\mathrm{a}}{\sqrt{\mathrm{v}^2-\mathrm{v}_1^2}} \end{aligned}

)

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