pre-cal help! [word problem]?

Question

the u.s.s. independence maintains a constant speed of 10 knots heading due north. at 4 pm the ship's radar detects a destroyer 100 nautical mi due east of the carrier. if the destroyer is heading due west at 20 knots, when will the 2 ships be the closest?
(1 knot = 1 nautical mi/hr)

thanks for the help.

Answer 1

9pm
First write down all the known.
Speed 1 = 10 knots north
distance from destroyer = 100 nautical mi east at 4pm
Speed 2 = 20 knots west

Now draw a picture


000000000000000000000000000|\
000000000000000000000000000|10 knots
100 nautical miles|
destroyer========|independence
-------->20 knots

distance at 4pm=100 knots east +0 knots north
distance at t=100 knots east+20 knots west(t) +10 knots north(t)
distance at t=100 knots east- 20 knots east(t) +10 knots north(t)
The min distance east is 0 nautical mi this occures at t =5h
Once the distroyer moves west more then a hunndred knots it gets further away.
At t=5 the distance in north=50 knots north at 4pm +5h= 9pm
At 10pm or t=6 the distance 20 knots west and 60 knots north

Answer 2

at 4 pm

Answer 3

at 10 pm.