Two-particles-A-and-B-move-with-constant-velocities-v-1-and-v-2-along-two-mutually-perpendicular-straight-lines-towards-the-intersection-point-O-At-moment-t-0-the-particles-were-located-at-dista

Question Number 19167 by Tinkutara last updated on 06/Aug/17

Two particles A and B move with

constant velocities v_{1} and v_{2} along two

mutually perpendicular straight lines

towards the intersection point O . At

moment t = 0, the particles were

located at distances d_{1} and d_{2} from O

respectively . Find the time, when they

are nearest and also this shortest

distance .

Commented by Tinkutara last updated on 06/Aug/17

Answered by ajfour last updated on 06/Aug/17

Commented by ajfour last updated on 06/Aug/17

v_{relative} = v_{r} = \sqrt{v_{1}^{2} + v_{2}^{2}}

AB = r = \sqrt{d_{1}^{2} + d_{2}^{2}}

α = (α + θ) - θ

rsin α = rsin (α + θ) \cos θ - rcos (α + θ) \sin θ

= d_{1} (\frac{v_{2}}{v_{r}}) - d_{2} (\frac{v_{1}}{v_{r}}) \dots . . (i)

rcos α = rcos (α + θ) \cos θ + \sin (α + θ) \sin θ

= d_{2} (\frac{v_{2}}{v_{r}}) + d_{1} (\frac{v_{1}}{v_{r}}) \dots . . (ii)

From figure,

shortest distance s is given by

s =∣ rsin α ∣

= \frac{∣ d_{1} v_{2} - d_{2} v_{1} ∣}{\sqrt{v_{2}^{2} + v_{1}^{2}}} [see (i)]

let time when at shortest distance

be t_{s},

v_{r} t_{s} = rcos α

t_{s} = \frac{rcos α}{v_{r}} = \frac{d_{1} v_{1} + d_{2} v_{2}}{v_{r}^{2}} [see (ii)]

t_{s} = \frac{v_{1} d_{1} + v_{2} d_{2}}{v_{1}^{2} + v_{2}^{2}} .

Commented by Tinkutara last updated on 06/Aug/17

Thank you very much Sir!