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January 30, 2015

January 30, 2015

Posted by **JEN** on Monday, July 15, 2013 at 9:04am.

- math -
**Steve**, Monday, July 15, 2013 at 10:22amWell, 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

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