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Ship A is 10km due west of ship B. Ship ...

Ship A is 10km due west of ship B. Ship A is heading directly north at a speed of 30 kmph while ship B is heading in a direction 60° west of north at a speed 20kmph. Their closest distance of approach will be........

Text Solution

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`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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