Question 130565: A patrol plane flies 220 miles per hour in still air. It carries fuel for 4 hours of safe flying. If it takes off on patrol against a wind of 20 miles per hour, how far can it fly and return safely?
Found 2 solutions by ptaylor, solver91311:
Answer by ptaylor(2198)   (Show Source):
You can put this solution on YOUR website!
Distance(d)=Rate(r) times Time(t) or d=rt; r=d/t and t=d/r
rate of the plane against the wind=220-20=200 mph
rate of the plane with the wind=220+20=240 mph
distance flown with the wind = distance flown against the wind
Time flying against the wind=d/200
Time flying with the wind=d/240
And we are told that these two times add up to 4 hours, so:
(d/200)+(d/240)=4 multiply each term by 4
d/50 + d/60=16
6d/300 + 5d/300=16
6d+5d=4800
11d=4800
d=436.36 mi---distance the plane can fly out and return safely (actually if I was the pilot I would shoot for about 420 mi out)
CK
436.36/200+436.36/240=4
3.99999999999999999999---~~4
Hope this helps---ptaylor
Answer by solver91311(24713)  (Show Source):
You can put this solution on YOUR website! I'm going to presume that the wind is constant for the entire 4 hour trip. That means that it will be a head wind for the outbound leg of the trip and a tail wind for the return trip.
If there were no wind, then the true speed and the speed through still air would be the same and the round trip would be  miles for a patrol range of 440 miles. One might think that if the outbound speed were reduced by 20 mph because of the wind, but the return leg speed were increased that same 20 mph, that everything would average out and you could still go the same 440 miles out and make it back.
That turns out not to be the case. If the outbound leg true speed is 200 (220 - 20) and the aircraft flies 440 miles, then the time taken on the outbound trip is  hours. On the return leg, that same 440 miles will take  hours -- and you will run out of gas 2 minutes (0.033... hours) before you get back.
So, let's analyze this a little more carefully.
We know that the distance out has to equal the distance back, and we know that the time out plus the time back has to equal 4 hours. We also know that the distance out is equal to 200 mph times the time out  and that the distance back is equal to 240 mph times the time back  . Putting this all together we come up with two equations:
 and 
Rearranging the second equation gives us  , so we can now substitute this expression for into the first equation:
Solving for :
 , or the repeating decimal 1.81...
is therefore  , or the repeating decimal 2.09...
So, if the aircraft travels at 200 mph for  hours, then the distance travelled is  or the repeating decimal 436.36...
As a check, the return distance must be  , answer checks.
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