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Question 21312: Okay i need to know how to solve this question, The U.S.S. Independence maintains a constant speed of 10 knots heading due north. At 4:00 pm the ship's radar detects a destroyer 100 nautical miles due east of the carrier. If the destroyer is heading due west at 20 knots, when will the two ships be the closet? (1 knot= 1 nautical mile)
Answer by Photonjohn(42)   (Show Source):
You can put this solution on YOUR website! You have a right triangle with northern leg of 10T or 10 knots for a time T.
You have the right leg of 100 miles minus 20 knots for a time T
So the distance between them is the hypotenuse which is the square of the sum of these two legs.
Distance D^2 = ((10T)^2 + (100-20T)^2
D = ( 100T^2 + (100-20T)^2)^1/2
differentiate this distance with respect to time T
dD/dT = d/DT( of the above)
set = to 0
let u = (...) above
0 = d/DT ( u )1/2 letting u = simplifed (...) = 500(T^2 - 8T + 20)
du/DT = 500(2T - 8)
yields
0 = 2T - 8
T = 4 hours
At 4 hours, the carrier is 40 nmiles north and the destroyer has moved 100 - 80 nmiles west. The distance is the square root of 40^2 + 20^2 or (2000)^1/2 = 44.7 miles apart.
checking out a time greater than 4 hours, say 5 hours, yields legs of 50 miles north and (100-100) east = square root of 2500 = 50 miles apart.
checking our a time less than 4 hours, say 3 hours, yields legs of 30 miles north and (100-60) = 40 miles east = square root of 2500 = 50 miles apart.
This shows a minimum at T = 4 hours which is 44.7 k miles apart.
If you have a graphing calculator, you can see the graph reach the minimum at 44.7 k miles.
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