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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, how far can it fly
and return safely? (Assume that the wind remains constant.)
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Let d be the maximum one way distance, in miles.
The rate of the plane (relative the ground) against the wind is 300-30 = 270 mils per hour.
The time to fly the distance of d miles is hours.
The rate of the plane (relative the ground) with the wind is 300+30 = 330 mils per hour.
The time to fly the distance of d miles is hours.
The total time flying is + .
The time equation is
+ = 5 hours.
To solve, multiply all the terms by 270*330. You will get
330d + 270d = 5*270*330
600d = 445500
d = 445500/600 = 742.5.
ANSWER. The maximum one-way distance to fly is 742.5 miles.
Solved.