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A plane flying the 3458-mi from New York City to London has a 40-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 to London.
If the plane's speed in still air is 840 mph, how far is New York
from the point of no return?
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Let x be the distance, in miles, from NY to the "point of no return".
Then you have this time equation
= .
Left side is the time to return to NY from point "x" flying against the wind;
right side is the time to get London from point "x" flying with the wind.
It is the same as
=
or
= .
Cross-multiply and simplify
11x = 10*(3458 - x),
11x = 34580 - 10x
11x + 10x = 34580
21x = 34580
x = 34580/21 = 4940/3 = 1646 miles = 1646.667 miles (rounded).
Solved.
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To better understand all auxiliary circumstances, you should know that the shortest avia-routes
are the arcs of great circles between starting and ending points at the planes through the center of the Earth -
they are not straight lines drawn on flat maps ( ! )
So, the wind during such flight not necessary follows that root along the great arc - it is
a pure mathematical assumption, which is made in this problem.