SOLUTION: Flying against the wind, an airplane travels 5580 in 9 hours. Flying with the wind, the same plane travels 4200 in 5 hours. What is the rate of the plane in still air and what is
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Question 331366: Flying against the wind, an airplane travels 5580 in 9 hours. Flying with the wind, the same plane travels 4200 in 5 hours. What is the rate of the plane in still air and what is the rate of the wind?
Found 2 solutions by Alan3354, solver91311:
Answer by Alan3354(69443) (Show Source): You can put this solution on YOUR website!
Flying against the wind, an airplane travels 5580 in 9 hours. Flying with the wind, the same plane travels 4200 in 5 hours. What is the rate of the plane in still air and what is the rate of the wind?
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5580 what? Furlongs? Miles? Nautical miles?
I'll assume nautical miles, since that's what aircraft use.
5580/9 = 620 knots
4200/5 = 840 knots
The airspeed is the average (620+840)/2 = 730 knots
The "speed in still air" is called the airspeed.
If the people writing the textbooks don't know that, they should not enter problems dealing with it.
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The wind speed is the difference between airspeed and ground speed
= 840 - 730
= 110 knots
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Aircraft use knots, not mph, for speeds, and so do the weather forecasters.
1 knot = 1 nautical mile per hour
Answer by solver91311(24713) (Show Source): You can put this solution on YOUR website!
You don't bother to mention the distance units that the plane travels. Not a real problem except that it makes rendering the units a bit awkward.
The rate of the plane relative to the ground when flying against the wind is 
where 
is the rate in still air and 
is the rate of the wind. Flying with the wind, 
We know that distance equals rate times time, so:
)
and
)
Putting the two equations into standard form:
Solve the system of equations for and
John
My calculator said it, I believe it, that settles it
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