Two particles, 1 and 2, move with constant velocities `v_1` and `v_2` along two mutually perpendicular straight lines toward the intersection point O. At the moment `t=0` the particles were located at the distances `l_1` and `l_2` from the point O. How soon will the distance between the particles become the smallest? What is it equal to?

Options

A. `|l_(1)v_(2)-l_(2)l_(1)|/sqrt(v_(1)^(2)+v_(2)^(2)),(l_(1)v_(1)+l_(2)l_(2))/(v_(1)^(2)+v_(2)^(2))`
B. `|l_(1)v_(1)-l_(2)l_(2)|/sqrt(v_(1)^(2)+v_(2)^(2)),(l_(1)v_(2)+l_(2)l_(1))/(v_(1)^(2)+v_(2)^(2))`
C. `|l_(1)v_(2)-l_(2)l_(1)|/sqrt(v_(1)^(2)+v_(2)^(2)),sqrt(l_(1)/l_(2)),((l_(1)v_(1)+l_(2)v_(2))l_(1))/((v_(1)^(2)+v_(2)^(2))l_(2))`
D. `|l_(1)v_(2)-l_(2)l_(1)|/sqrt(v_(1)^(2)+v_(2)^(2)),sqrt(l_(2)/l_(1)),((l_(1)v_(1)+l_(2)v_(2))l_(2))/((v_(1)^(2)+v_(2)^(2))l_(1))`

Answer

A

Explanation

Let the separation between the particle be minimum at time `t`,Then Since `OB=l_(2)-v_(2)t` and `OA=l_(1)-v_(1)t` and `AB^(2)=OB^(2)+OA^(2)rArr s^(2)=(l_(1)-v_(1)t)^(2)+ (l_(2)-v_(2)t)^(2)`
For `s` to be minimum `(ds)/(dt)=0` or `(ds)/(dt)(s^(2))=0`
`rArr 2s(ds)/dt=2(l_(1)-v_(1)t)-v_(1)+2(l_(2)-v_(2)t)-v_(2)=0`
`-l_(1)v_(1)+v_(1)^(2)t-l_(2)^(2)+v_(2)^(2)t=0rArrt=(l_(1)v_(1)+l_(v)v_(2))/(v_(1)^(2)+v_(2)^(2))`
`S_min^(2) =[l-((l_(1)v_(1)+l_(2)v_(2))/(v_(1)^(2)+v_(2)^(2)))v_(1)]^(2)+[l_(2)-v_(2)((l_(1)v_(1)+l_(2)v_(2))/(v_(1)^(2)+v_(2)^(2)))v_(1)]^(2)`
`S_ min^(2)=(l_(1)v_(2)-l_(2)v_(1))^(2)/(v_(1)^(2)+v_(2)^(2))rArr S_(min)=(|l_(1)v_(2)-l_(2)v_(1)|)/sqrt((v_(1)^(2)+v_(2)^(2)))`