SOLUTION: 3. An airplane flew 4 hours with a 25 mph tail wind. The return trip against the same wind took 5 hours. Find the speed of the airplane in still air. This similar to the current pr

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Question 57327: 3. An airplane flew 4 hours with a 25 mph tail wind. The return trip against the same wind took 5 hours. Find the speed of the airplane in still air. This similar to the current problem as you have to consider the 25 mph tailwind and headwind.
Let r = the rate or speed of the airplane in still air.
Let d = the distance
Write a system of equations for the airplane. One equation will be for the outbound trip with tailwind of 25 mph. The second equation will be for the return trip with headwind of 25 mph. Solve the system of equations for the speed of the airplane in still air.

Answer by stanbon(75887) (Show Source):
You can put this solution on YOUR website!
. An airplane flew 4 hours with a 25 mph tail wind. The return trip against the same wind took 5 hours. Find the speed of the airplane in still air. This similar to the current problem as you have to consider the 25 mph tailwind and headwind.
Let r = the rate or speed of the airplane in still air.
Let d = the distance
Write a system of equations for the airplane. One equation will be for the outbound trip with tailwind of 25 mph. The second equation will be for the return trip with headwind of 25 mph. Solve the system of equations for the speed of the airplane in still air.
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With the wind DATA:
time= 4 hrs ; rate = (r+25) mph ; distance = 4(r+25)
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Against the wind DATA:
time=5 hrs.; rate = (r-25) mph ; distance = 5(r-25)
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EQUATION:
1st: d=4(r+25)
2nd: d=5(r-25)
Substitute to get:
4(r+25)=5(r-25)
4r+100=5r-125
r=225 mph (speed of the airplane in still air)
Substitute into 1st to solve for "d"
d=4(225+25)
d=4(250)
distance=1000 miles
Cheers,
Stan H.