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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=vAvB
vAB=(20 ^i20 ^j) km/h
Hence |vAB|=400+400=202 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=90θ=45
Hence shortest distance between ships is BP,
sinα=BPAB
12=BP50
BP=502=252 km
For ship B, the ship A appears to be coming towards it along line AP.
From ΔBAP,
AP=AB2BP2=25001250
AP=252 km
|vAB|×t=AP
t=AP|vAB|=252202=1.25 h
t is the time at which shortest distance occured.

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