SOLUTION: A plane flying the 3458 mi. trip from New York City to London has a 50 mph tailwind. The flight's point of no return is the point at which the flight time required to return to Ne

Question 148457This question is from textbook
: A plane flying the 3458 mi. trip from New York City to London has a 50 mph tailwind. The flight's point of no return is the point at which the flight time required to return to New York is the same as the time required to continue on to London. If the speed of the plane in still air is 360 mph, how far is New York from the point of no return. This question is from textbook

Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
A plane flying the 3458 mi. trip from New York City to London has a 50 mph tailwind.
The flight's point of no return is the point at which the flight time required to return to New York is the same as the time required to continue on to London.
If the speed of the plane in still air is 360 mph, how far is New York from the point of no return.
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With the wind data:
rate = 360+50 = 410mph ; distance = x mi ; time = d/r = x/410
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Again the wind data:
rate = 360-50 =310 mph ; distance = (3458-x) mi ; time = d/r = (3458-x)/310
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EQUATION:
After going x miles from NY, time to go the remaining miles with the wind
equals time to return x miles against the wind.
time with = time against
x/310 = (3458-x)/410
410x = 310*3458 - 310x
720x = 310*3458
x = 1488.86 miles from NY is the point of no return
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Cheers,
Stan H.

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