SOLUTION: A plane flying the 3020-mile trip from City A to City B has a 60-mph tailwind. The flight's point of no return is the point at which the flight time required to return to City A

Question 279305: A plane flying the 3020-mile trip from City A to City B has a 60-mph tailwind. The flight's point of no return is the point at which the flight time required to return to City A is the same as the time required to continue to City B. If the speed of the plan in still air is 430 mph, how far from City A is 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 3020-mile trip from City A to City B has a 60-mph tailwind.
The flight's point of no return is the point at which the flight time required
to return to City A is the same as the time required to continue to City B.
If the speed of the plan in still air is 430 mph, how far from City A is the
point of no return?
:
Determine the speeds
430+60 = 490 mph toward city B (with the wind)
430-60 = 370 mph back to city A (against the wind)
:
Let d = distance from City A to the point of no return
then
(3020-d) = distance from city B to this point
:
At this point the travel time to each city will be the same
Write a time equation; Time = dist/speed
:
Time to return against the wind = time to continue with the wind
=
Cross multiply
490d = 370(3020-d)
:
490d = 1117400 - 370d
:
490d + 370d = 1117400
:
860d = 1117400
d =
d = 1299.3 miles from A
:
:
You can prove this: 3020 - 1299.3 = 1720.7 to city B
Time back to A: 1299.3/370 = 3.511 hrs
Time on to B: 1720.7/490 = 3.511 hrs also
:
Did this make sense to you?

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