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

A. a/√v²+v1²
B. √a²/(v²−v1²)
C. a/√v−v1
D. a/(v+v1)

Answer 1

The correct expression for the time 't' it takes the boy at A to catch the boy at B, given their velocities and starting positions, is t = a/√(v^2 - v1^2), assuming v > v1. The question's provided options contained a typo, with the correct answer being closest to Option B after correcting the square root's term.

Step-by-step explanation:

The problem describes a scenario with two boys running, one perpendicular to the line that joins them and the other directly towards him. To determine the time it takes for the second boy to catch the first, we can use the relative velocity concept in two dimensions. The boy at B runs with velocity v1 perpendicularly to AB, while the boy at A chases with velocity v along the line AB. The time t it takes for the boy at A to catch the boy at B can be found by using Pythagoras' theorem, considering the distance covered by each boy and their relative motion.

To find the time t, we can create a right-angled triangle with sides of length a (the initial distance between the boys), vt (the distance the boy at A would run along AB in time t) and v1t (the distance the boy at B would run perpendicular to AB in time t). The hypotenuse would represent the path of the boy at A as he catches up to the boy at B. Using the Pythagorean theorem:

a2 + (v1t)2 = (vt)2,

we solve for t and get t = a/√(v2 - v12), which is not one of the options provided in the original question. If we consider that boy A must run with a velocity greater than that of boy B (v > v1) to catch him, then the answer is Option B. It seems the original options had a typo, and I must rectify that in my response.