posted by UCI Student on .
At noon, ship A is 40 nautical miles due west of ship B. Ship A is sailing west at 19 knots and ship B is sailing north at 20 knots. How fast (in knots) is the distance between the ships changing at 7 PM?
i really don't have any idea what to do...
The relative velocity vector is
Vab = Va - Vb = -19*t i + 20t j
where i is a unit vector east and j is a unit vector north.
The magnitude of that vector is the rate of change of the distance between them, and is
|Vab| = sqrt[19^2 + 20^2] = 27.59 knots
That rate is a constant, and does not depend upon time of day nor the initital 40 n.m. separation. You have been given more information than you need.
The actual separation distance at any time DOES depend upon time and initial separation.