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Topic: THE CHASE (Read 551 times) |
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pcbouhid
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Ship A is chasing ship B and making 30 knots to 15 for the pursued, and neither is equipped with radar.
Ship B enters a cloud bank which makes its further visibility from ship A impossible.
Captain A correctly assume that Captain B will take advantage of the fog to immediately change his course and will mantain his new direction unchanged at full speed.
Based on this assumption, what plan should Captain A follow to insure that he will intercept ship B?
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| « Last Edit: Dec 2nd, 2005, 8:25am by pcbouhid » |
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Joe Fendel
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Re: THE CHASE
« Reply #1 on: Dec 2nd, 2005, 12:52pm » |
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I may get really seasick  , but I'll try this anyway.
We're not given the shape of the cloud, but let's assume the "worst case" scenario, that the cloud is spreading in every direction faster than 15 knots, so that Captain B has full 360-degree freedom with his angle choice.
The set of possible locations of B can be described by a ripple moving outward from the point when B entered the cloud. If (theta, x) are the polar coordinates associated with B's location with respect to this point, then we have x = 15t and theta = thetaB, a constant.
Thus A should continue his chase until his boat meets this ripple, at time t0. At this point, A begins searching around the ripple, by following a route given by polar coordinates (thetaA(t), 15t).
What is left is to find the function thetaA(t). My calculus is a bit rusty, but I think the speed of A's boat can be shown to be equal to 15*sqrt(1 + (t * thetaA'(t))^2), which we know is 30. Thus thetaA'(t) = sqrt(3) / t, so thetaA(t) = sqrt(3)*ln(t) - C, where C = sqrt(3)*ln(t0).
Since the natural log function increases unboundedly, A will eventually traverse the whole ripple and catch B.
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