Question

Two boys are standing at the ends A and B of a ground where AB = a . The boy at B starts running in direction pependicular to AB with a velocity v1 . The boy at A starts running simultaneously with velocity v and catches the other boy in a time t, where t is

(1) a/ (v2 + v12)1/2

(2) a/ (v + v1)

(3) a/(v - v1)

(4) ( a2/ (v2- v12)1/2

Answer 1

Hey there! ☺☻☺

Given,
Two boys standing at the Ends A & B of a ground.
The Boy at B start running in direction perpendicular to AB. With velocity v1.
The boy at A start running simultaneously with velocity v and catches other boy in time t.
Let the two boys meet at C in t time.

Hence, AC = vt 
            BC = v1t
By Pythagoras,
$AC^2 = BC^2 + AB^2\\v^2t^2 = v^2_1t^2 + a^2\\\\\boxed{\boxed{\bold{t = \sqrt{\frac{a^2}{v^2 - v^2_1}}}}}$

Hope It Helps You! ☺☻☺

Answer 2

Explanation:

the complete answer is there in the image.