SOLUTION: A plane flies into the wind (against the wind current) for 3.5 hours for a distance of 1,260 miles. The return trip with a tailwind (with the wind current) only takes 3 hours. Find

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Question 1127565: A plane flies into the wind (against the wind current) for 3.5 hours for a distance of 1,260 miles. The return trip with a tailwind (with the wind current) only takes 3 hours. Find the average speed of the plane in still air and the speed of the wind current.
Found 2 solutions by josgarithmetic, MathTherapy:
Answer by josgarithmetic(39799) About Me  (Show Source):
You can put this solution on YOUR website!
system%28r-w=1260%2F3.5%2Cr%2Bw=1260%2F3%29

-

w%2B1260%2F3.5=-w%2B1260%2F3
2w=1260%2F3-1260%2F3.5
2w=60
highlight%28w=30%29---------speed of wind

highlight%28390=r%29--------speed of plane in absence of wind

Answer by MathTherapy(10810) About Me  (Show Source):
You can put this solution on YOUR website!

A plane flies into the wind (against the wind current) for 3.5 hours for a distance of 1,260 miles. The return trip with a tailwind (with the wind current) only takes 3 hours. Find the average speed of the plane in still air and the speed of the wind current.
Let the speed of the plane, in still air, and the speed of the wind, be S and W, respectively

Then we get the following TIME equations: 

2S = 780 ------- Adding eqs (ii) & (i)

S, or speed, in still air = highlight_green%28matrix%281%2C4%2C+780%2F2%2C+%22=%22%2C+390%2C+mph%29%29

W, or windspeed: highlight_green%28matrix%281%2C4%2C+420+-+390%2C+%22=%22%2C+30%2C+mph%29%29