SOLUTION: An air rescue plane averages 300 miles per hour in still air. It carries enough fuel for 5 hours of flying time. If, upon takeoff it encounters a head wind of 30 mi/hr and the wi

Question 268621: An air rescue plane averages 300 miles per hour in still air. It carries enough fuel for 5 hours of flying time. If, upon takeoff it encounters a head wind of 30 mi/hr and the wind remains constant, how far can the plane fly and then return safely?
Is the rate simply 300-30? How do I find the distance traveled?
Found 2 solutions by ankor@dixie-net.com, dabanfield:
Answer by ankor@dixie-net.com(22740) (Show Source): You can put this solution on YOUR website!
An air rescue plane averages 300 miles per hour in still air.
It carries enough fuel for 5 hours of flying time.
If, upon takeoff it encounters a head wind of 30 mi/hr and the wind remains constant, how far can the plane fly and then return safely?
:
Effective ground speed
300 - 30 = 270 mph against the wind
300 + 30 = 330 mph with the wind
:
Let d = the one-way distance; write a time equation, time = dist/speed
:
outbound time + return time = 5 hrs
+ = 5
Multiply by 81900 (270*330), eliminate the denominators, results:
330d + 270d = 5(81900)
600d = 445500
6d = 4455
d =
d = 742.5 mi, the plane can fly out and return safely
:
:
Check this by finding the total time
742.5/270 + 742.5/330 =
2.75 + 2.25 = 5

Answer by dabanfield(803) (Show Source): You can put this solution on YOUR website!
An air rescue plane averages 300 miles per hour in still air. It carries enough fuel for 5 hours of flying time. If, upon takeoff it encounters a head wind of 30 mi/hr and the wind remains constant, how far can the plane fly and then return safely?
Is the rate simply 300-30? How do I find the distance traveled?
The time the plane can stay in the air is 5 hours. As you determined, the speed that the plane can fly against the wind is 300-30. But on a round trip the plane will fly back with a speed of 300+30.
Let t be the time taken for the trip against the wind. Then the part of the trip with the wind will be 5-t since the whole trip will take 5 hours. Let's say the distance for each leg of the trip is d. Then we have:
1.) d = rate*time = (300-30)*t (against the wind)
2.) d = rate*time = (300+30)*(5-t) (with the wind)
Since both expressions are equal to d we have:
270*t = 330*(5-t)
Solve the above for t then calculate the distance d by substituting this value for t in either 1.) or 2.) above.
Remember the TOTAL distance traveled is 2*d.

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