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 is

Algebra ->  Coordinate Systems and Linear Equations  -> Linear Equations and Systems Word Problems -> 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 is      Log On


   

Question 260649: 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 plane in still air is 430 mph, how far from City A is the point of no return? Round your answer to the nearest mile.
Answer by ankor@dixie-net.com(22740) About Me  (Show Source):
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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 plane in still air is 430 mph, how far from City A is the point of no return? Round your answer to the nearest mile.
:
Let x = distance from city A
Then
(3020-x) = distance from city B
:
Write a time equation: time = dist/speed
:
Continue time = return time
%28%283020-x%29%29%2F%28%28430%2B60%29%29 = x%2F%28%28430-60%29%29
:
%28%283020-x%29%29%2F490 = x%2F370
:
Cross multiply
490x = 370(3020-x)
490x = 1117400 - 370x
490x + 370x = 1117400
860x = 1117400
x = 1117400%2F860
x = 1299 mi from A is the point of no return
:
:
Check solution by finding the times
1299/370 = 3.5 hrs to return to A
(3020-1299)/490 = 3.5 hrs to continue to B