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Ship A is 10km due west of ship B. Ship A is heading directly north ar a speed of 30 kmph while ship B is heading in a direction 600 west of north at a speed 20kmph. Their closest distance of approach will be.... |
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Answer» Solution :`bar(V_(A))=30 j bar(V_(B))=20 sin 60^(@)(i)+20 Cos(60)(HATJ)` `=-10sqrt(3i)+10 hatj` `bar(V_(BA))=bar(V_(A))=-20hati- 10 SQRT(3i)` If `phi` is the angle MADE by `bar(V_(BA))` with X axis `TAN phi=(20)/(10 sqrt(3))=(2)/(sqrt(3)) and "sin" phi=(2)/(sqrt(7))` From `DeltaABC,(x)/(10)XX(2)/( sqrt(7))` `x=(20)/(sqrt(7))=7.6 Km` 
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