Question 1005093: Please help me set up this question: A plane that can fly 200 mph in still air makes a 330 mile flight with a tail wind and returns flying into the same wind. Find the speed of the wind if the total flying time is 3 1/3 hours.
Answer by Theo(13342)   (Show Source):
You can put this solution on YOUR website! speed of the plane = 200 miles per hour.
let w = the speed of the wind.
with the wind, the combined speed of the plane and the wind is 200 + w.
against the wind, the combined speed of the plane and the wind is 200 - w.
the plane travels 330 miles with the wind and 330 miles against the wind.
total flying time is 3 and 1/3 hours = 10/3 hours.
let T1 = the amount of time it takes flying with the wind.
let T2 = the amount of time it takes flying against the wind.
the rate * time = distance equations become:
(200 + w) * T1 = 330
(200 - w) * T2 = 330
solve for T1 and T2 to get:
T1 = 330 / (200 + w)
T2 = 330 / (200 - w)
since T1 + T2 is equal to 3 and 1/3 hours = 10/3 hours, we get:
T1 + T2 = 10/3 hours.
after your replace T1 and T2 with their equivalents, the equation becomes:
330 / (200 + w) + 330 / (200 - w) = 10/3
multiply both sides of this equation by 3 and divide both sides of this equation by 10 to get:
99 / (200 + w) + 99 / (200 - w) = 1
multiply both sides of this equation by (200 + w) * (200 - w) to get:
99 * (200 - w) + 99 * (200 + w) = (200 + w) * (200 - w)
the algebra gets a little ugly, but, when you solve this equation for w, you get:
w = 20 miles per hour.
you can confirm this solution is correct by replacing w in the original equations to see what you get:
the original equations to work with are:
T1 = 330 / (200 + w)
T2 = 330 / (200 - w)
solving for T1 and T2, using w = 20 miles per hour.
T1 = 1.5 hours
T2 = 1.833333 hours
T1 + T2 = 3.33333333..... which is the same as 3 and 1/3 which is the same as 10/3.
if you need help with the ugly, let me know.
otherwise i'll assume you can handle it.
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