SOLUTION: a plane flying the 4258-mi trip from new york to london has a 50-mph tailwind. The flights point of no return is the point at which the flight time required to return to new york i
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Question 598568: a plane flying the 4258-mi trip from new york to london has a 50-mph tailwind. The flights 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 speed of the plane in still air is 360 mph, how far is new york from the point of no return? Answer by ankor@dixie-net.com(22740) (Show Source):
You can put this solution on YOUR website! a plane flying the 4258-mi trip from new york to London has a 50-mph tailwind.
The flights 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 speed of the plane in still air is 360 mph, how far is new york from the point of no return?
:
Let d = distance from NY to point of no return
Then
(4258-d) = distance of that point from London
:
360 + 50 = 410 mph effective speed with the wind, (going east)
and
360 - 50 = 310 mph effective speed against the wind, (going west)
:
Time to return to NY = Time to continue to London =
cross multiply
410d = 310(4258-d)
410d = 1319980 - 310d
310d + 410d = 1319980
720d = 1319980
d = 1319980/720
d = 1833 mi from New York
:
:
Check this by finding the flight times from d to London and from d back to NY
(4258-1833)/410 = 5.9 hrs
1859/310 = 5.9 hr
