Well, since the ship is equidistant from A and B, it lies on the perpendicular bisector of AB, so if A is at (0,0) and B is at (6,0), the ship is at (3,4).
If the ship is at (x,y), the distances u from A and v from B are
u^2 = x^2 + y^2
v^2 = (x-6)^2 + y^2
2u du/dt = 2x dx/dt + 2y dy/dt
2v dv/dt = 2(x-6) dx/dt + 2y dy/dt
When x is at (3,4)
3 dx/dt + 4 dy/dt = 140
-3 dx/dt + 4 dv/dt = 20
dx/dt = 20
dy/dt = 20
Looks like it's moving NE at 20√2 km/h
Two radar stations at A and B, with B 6 km east of A, are tracking a ship which is generally to the north. At a certain instant, the ship is 5 km from A and this distance is increasing at the rate of 28 km/h. At the same instant, the ship is also 5 km from B, but this distance is increasing at only 4 km/h. Where is the ship, how fast is it moving, and in what direction is it moving?
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