Question

# A ship A is moving eastwards with a speed of 20 km/h and a ship B ,50 km south of A is moving northwards with a speed of 20 km/h. The time after which the distance between ships become shortest is

A
1 h
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B
3.5 h
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C
1.25 h
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D
2.5 h
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Solution

## The correct option is C 1.25 h−→vA=20^i km/h −→vB=20^j km/h velocity of A w.r.t B is, −−→vAB=−→vA−−→vB −−→vAB=(20 ^i−20 ^j) km/h Hence |−−→vAB|=√400+400=20√2 km/h ∴ tanθ=|−20|20=1 ⇒θ=45∘ ∵ velocity of A w.r.t B is directed along direction AP, so the minimum distance (dmin) occurs where relative velocity of A w.r.t B i.e −−→vAB is ⊥ to B. ⇒ From ΔAPB, α=90∘−θ=45∘ Hence shortest distance between ships is BP, sinα=BPAB ⇒1√2=BP50 ∴BP=50√2=25√2 km For ship B, the ship A appears to be coming towards it along line AP. From ΔBAP, AP=√AB2−BP2=√2500−1250 AP=25√2 km ⇒ |−−→vAB|×t=AP ∴t=AP|−−→vAB|=25√220√2=1.25 h ⇒ t is the time at which shortest distance occured.

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