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A plane flew 600 miles with a 30mph tailwind. On the return trip, the plane flew 600 miles into a 30mph headwind. If the total flying time was 4.5 hours, how fast does the plane fly in still air? PLEASE HELP!
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x = rate in still air
x + 30 = rate with wind x - 30 = rate against wind D/R = T 600/(x + 30) = Time with wind 600/(x - 30) = time against wind 600/(x+30) + 600/(x - 30) = 4.5 total hours or 9/2 multiply by LCD= 2(x+30)(x - 30) 1200(x - 30) + 1200(x + 30) = 4.5*2(x+30)(x -30) 1200x - 3600 + 1200x + 3600 = 9(x^2 - 900) 2400x = 9x^2 - 8100 0 = 9x^2 -2400x -8100 x= 270 or -3.3 270 mph in still air |
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The answer is 267 mph
Neglect the 30mph wind coming and going because they will cancel. So 1200 miles / 4.5 hours equals 267 mph The plane will go 297 mph and will take 2 hours to go 600 miles The plane coming will go 237 mph and will take 2.5 hours to go 600 miles. 2 hours + 2.5 hours add up to 4.5 hours. |
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Neil, they will not cancel. Think about it. The headwind slows the airplane down causing it to stay enroute to its destination longer thus it has more time than the tailwind, which conversely by speeding the airplane has less time, to effect the average speed. NTSB reports have plenty of account of pilots budgeting enough fuel for a return trip after they neglected to take winds into account because they felt the headwind would cancel the tailwind.
270 is indeed the correct answer. |
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