SOLUTION: Any help on this problem would be greatly appreciated! Thank you in advance!
An air rescue plane averages 300 miles per hour in still air. It carries enough fuel for 5 hours of
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Question 137397: Any help on this problem would be greatly appreciated! Thank you in advance!
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 wind of 30 miles per hour, how far can it fly and return safely? ( assume that the wind remains constant).
Thanks to anyone who can provide help on this one!
Answer by dolly(163) (Show Source): You can put this solution on YOUR website!
Hi,
Let the distance covered be = x miles.
Speed of the plane = 300 mph
Speed of wind = 30 mph
So relative speed while going against the wind = 300 - 30 = 270 mph
Thus time taken for covering x miles = distance/rate
= x/270 hrs
relative speed while retuning alongwith the wind = 300 + 30 = 330 mph
Thus time taken for returning x miles = distance/rate
= x/330 hrs
Given the total time taken = 5 hrs
Thus x/270 + x/330 = 5
==> = 5
[taking 2970 as the common denominator]
==> = 5
==> 20x = 5 * 2970
==> x = 742.5
==> one way the distance = 742.5 miles
So 2 way the distance = 1485 miles.
Thus the plane can safely fly 1485 miles and return with its fuel availability.
Good luck!!!
dolly
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