Question

Two particles A and B are initially at distance d apart. They start moving simultaneously at velocity v and u such that A always move towards B and B moves along fixed straight line which is perpendicular to initial direction of motion of A. Then particles will meet after time
A. (vd)/(v^2-u^2)
B. (ud)/(u^2-v^2)
C. d/(v^2+u^2)^1/2
D. vd/(v^2+u^2)

Answer 1

Let after time "t" the Position of A and B are in such a way that A is heading to B and its velocity makes some angle with x axis

So the relative speed of A with respect to B in horizontal direction is given as

$\frac{dx}{dt} = u - vcos\theta$

relative motion in y direction is given as

$\frac{dy}{dt} = 0 - vsin\theta$

from first equation we can say

$\int dx = \int u dt - \int vcos\theta dt$

$0 = uT - v\int cos\theta dt$

$\int cos\theta dt = \frac{uT}{v}$

now in the direction of approach of each other

$\frac{dr}{dt} = ucos\theta - v$

$\int dr = \int ucos\theta dt - \int vdt$

$0 - d = u\int cos\theta dt - vT$

$- d = u(\frac{uT}{v}) - vT$

$T = \frac{vd}{v^2 - u^2}$