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Two particles A and B are located at points `(0, -10sqrt3)` and (0, 0) in `x-y` plane. They start moving simultaneously at the time `t=0s` with constant velocities `vec(V_(A))=5hati" m s"^(-1) and vec(V_(B))=-5 sqrt3hatj" m s"^(-1)`, respectively. The time when they are closest to each other is found to be `K//2` second. Find K. All distances are given in meter.

Text Solution

Verified by Experts

The correct Answer is:
`00003.00`
`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
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