Two particles `A` and `B` located at points `(0,-10sqrt(3))` and `(0,0)` in `xy` plane. They start moving simultaneously at time `t=0` with constant velocities `vecV_(A)=5hatim//s` and `vecV_(B)=-5sqrt(3)hatjm//s`, respectively. Time when they are closest to each other is found to be `K//2` second. Find `K`. All distance are given in meter.

Answer

`00003.00`

Explanation

`00003.00`
As observed by `B` motion
of `A` observed by `B` motion
of `A` is along `AM` and `BM`
is the shortest distance
between them. Relative displacement of `A` w.r.t to `B`
Time taken `(t)_=|(vecS_(AB))/(vecV_(AB))|=(AM)/(V_(AB))=(10sqrt(3)cos30^(@))/10=1.5` sec