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timama [110]
4 years ago
6

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)
Physics
1 answer:
lozanna [386]4 years ago
5 0

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}

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Based on the above information

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