Question 130565
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 {{{4*220=880}}} 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 {{{440/200=2.2}}} hours.  On the return leg, that same 440 miles will take {{{440/240=1.8333}}} 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 {{{t[o]}}} and that the distance back is equal to 240 mph times the time back {{{t[b]}}}.  Putting this all together we come up with two equations:


{{{240t[b]=200t[o]}}} and {{{t[o]+t[b]=4}}}


Rearranging the second equation gives us {{{t[o]=4-t[b]}}}, so we can now substitute this expression for {{{t[o]}}} into the first equation:


{{{240t[b]=200(4-t[b])}}}


Solving for {{{t[b]}}}:


{{{240t[b]+200t[b]=800}}}


{{{440t[b]=800}}}


{{{t[b]=20/11}}}, or the repeating decimal 1.81...


{{{t[o]}}} is therefore {{{4-20/11=24/11}}}, or the repeating decimal 2.09...


So, if the aircraft travels at 200 mph for {{{24/11}}} hours, then the distance travelled is {{{200(24/11)=4800/11}}} or the repeating decimal 436.36...


As a check, the return distance must be {{{240(20/11)=4800/11}}}, answer checks.