Question
Question: Two boys are standing at the ends A and B of a ground, where \[AB = a\]. The boy at B starts running...
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) v2+v12a
B) v2−v12a2
C) v+v1a
D) v−v1a
Solution
First we see that according to the question, at which direction the boy B is moving. Then draw a figure to calculate time t at which Boy at A catches boy which runs from B. We use Pythagoras’s theorem to calculate time. According to the question, both boys are running along the sides of the right angle triangle.
Complete step by step solution:
First we draw diagram indicating directions of velocities of boys moving from A and B
Given: Width of ground, AB=a, velocity of boy moving from point B=v1, velocity of boy moving from point A=v
From figure, directions ofv1, v and AB are making sides of right angle triangle
Use Pythagoras theorem, AC2=AB2+BC2......(I)
Distance AC is given by, AC=velocity of boy moving from B ×time taken from A to C
⇒AC=vt
Distance AB is given by AB=a
Distance BC is given by, BC=v1t
Substituting values of AB,BC and CA in equation (i), we get
(vt)2=(a)2+(v1t)2