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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!

2007-12-06 06:15:02 · 3 answers · asked by Scott 1 in Science & Mathematics Mathematics

3 answers

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

2007-12-06 06:21:54 · answer #1 · answered by Linda K 5 · 0 0

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.

2007-12-09 10:01:45 · answer #2 · answered by Kevin 5 · 0 0

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.

2007-12-06 06:51:21 · answer #3 · answered by Neil 7 · 0 0

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