SOLUTION: An air rescue plane averages 300 miles per hour in still air. It carries fuel for 5 hours of flying time. If, upon takeoff, it encounters a head wind of 30 miles per hour, how far

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 Question 531211: An air rescue plane averages 300 miles per hour in still air. It carries fuel for 5 hours of flying time. If, upon takeoff, it encounters a head wind of 30 miles per hour, how far can it fly and return safely? (Assume that the wind remains constant)Found 2 solutions by ankor@dixie-net.com, Alan3354:Answer by ankor@dixie-net.com(15656)   (Show Source): You can put this solution on YOUR website!An air rescue plane averages 300 miles per hour in still air. It carries fuel for 5 hours of flying time. If, upon takeoff, it encounters a head wind of 30 miles per hour, how far can it fly and return safely? (Assume that the wind remains constant) : 300 - 30 = 270 mph the ground speed against the wind 300 + 30 = 330 mph the ground speed with the wind : let d = the one way distance to point of no return : Write a time equation, time = dist/speed : Outbound time + return time = 5 hrs + = 5 multiply by 2970 to get rid of the denominators, results 11d + 9d = 2970(5) 20d = 14850 d = d = 742.5 mi, but it says "safely", you would never cut it that close; I would say 650 max. : : Confirm this by finding the actual time for each trip 742.5/270 = 2.75 hrs 742.5/330 = 2.25 hrs --------------------- totals time: 5 hrs Answer by Alan3354(30993)   (Show Source): You can put this solution on YOUR website!An air rescue plane averages 300 miles per hour in still air. It carries fuel for 5 hours of flying time. If, upon takeoff, it encounters a head wind of 30 miles per hour, how far can it fly and return safely? (Assume that the wind remains constant) -------------------- The plane's avg groundspeed for the round trip is 2*330*270/(330 + 270) = 297 mi/hr Round trip distance = 5*297 = 1485 miles One way = 742.5 miles