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31 Oct 2009 12:05:45 IST

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Two ships are 10 km apart on a line running south to north. The one farther north is streaming west

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Two ships are 10 km apart on a line running south to north. The one farther north is streaming west at 20 kmh-1. The other is streaming north at 20 kmh-1. What is the distance of closest approach.

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shashank agarwal

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31 Oct 2009 12:14:05 IST

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Suppose the two ships X, Y moving with velocities u, v respectively each 20 km h^-1. The velocity of Y relative to X = v - u = v + (-u)

We therefore draw OA to represent v and add to it AB which represents -u. The relative velocity is then represented by OB.

OB = √(OA² + AB²) = √(20² + 20²)= 28.28 km h^-1.

Also tanθ = AB/OA = 20/20 = 1 => θ = 45^o

Thus the ship Y will be move along a direction QR relative to the ship X where QR is at 45^o to PQ, the north south direction. When the relative velocity is considered, the ship X is at rest

rate if u like

If PQ = 10 km the distance of closest approach is PN where PN is the perpendicular from P to QR.

PN = PQ sin 45^o = 10 sin 45^o = 7.71 km

The distance QN = 10 cos 45^o = 8.071 km

Time to reach N = 7.071/28.28 = 0.25h

VARUN RAJ

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31 Oct 2009 21:38:19 IST

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u can also solve it by normal calculus

Krishna Gopal Singh

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Joined: 29 Dec 2006

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31 Oct 2009 23:03:17 IST

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Right answer. But here is another approch.

Let us consider east as x-axis and north as y-axis. If initial positions of ships are (0,10) and (0,0) then there position at any time t is (-10t,10) and(0,10t)

Distance at any time t = sqrt(200t^2 -200t+100)

Differentiating we get that this happens at t = 0.5 hr (shashank please check why you are getting 0.25hr)

So distance = sqrt(50) = 7.07 km

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