You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XII

Physics

Relative 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

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is C.

Let two boys meet at point C after time 't' from the starting.Then $A C = v t, B C = v_{1} t$

Therefore in a right angle triangle ABC,
$(A C)^{2} = (A B)^{2} + (B C)^{2} \Rightarrow v^{2} t^{2} = a^{2} + v_{1}^{2} t^{2}$
$v^{2} t^{2} - v_{1}^{2} t^{2} = a^{2}$
$t^{2} (v^{2} - v_{1}^{2}) = a^{2}$
$\Rightarrow t = \sqrt{\frac{a^{2}}{v^{2} - v_{1}^{2}}}$ .

Suggest Corrections

138

Similar questions

Q. Two boys are standing at the ends

A

and

B

of a ground, where

A B = a

. The boy at

B

starts running in a direction perpendicular to

A B

with velocity

v_{1}

. The boy at

A

starts running simultaneously along a straight line with velocity

v

and catches the other boy in a time

t

, where t is

Q. 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

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

A B = a

. The boy at B starts running in a direction perpendicular to

A B

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

A boy runs along a straight path for the first half of the distance with a velocity $v_{1}$ and the second half of the distance with a velocity $v_{2}$ . The average velocity V is given by

Q. A boy runs along a straight path for the first half distance with a velocity

v_{1}

and the second half distance with a velocity

v_{2}

. The mean velocity V is given by

Join BYJU'S Learning Program

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

The correct option is C. Let two boys meet at point C after time 't' from the starting.Then AC=vt, BC=v1t Therefore in a right angle triangle ABC, (AC)2=(AB)2+(BC)2⇒v2t2=a2+v21t2v2t2−v21t2=a2t2(v2−v21)=a2⇒ t=√a2v2−v21.

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