Question 199712: An airplane flew for 4 hours against a head wind of 40 km/h. On the return flight the same wind was a tail wind, and the flight took 3 hours. Find the speed of the airplane in still air.
Answer by tutorvolunteer(9)   (Show Source):
You can put this solution on YOUR website! Since distance = speed x time, time = speed/distance +or- wind speed/distance
(+or- depending on whether it is in the same direction as plane, in which case it would be +, or opposite direction and -)
time = (speed +or- wind speed)/distance
(abbreviations used: time: t, speed in still air: s, wind speed: ws, distance: d)
Let's look at the first scenario: t = 4, ws = -40. Therefore, since t = (s - ws)/d, we get this equation:
4 = (s - 40)/d ----) this has 2 variable, which means we will need a system of equations to solve it.
Let's look at the second part of the question to get the second equation. Since t=(s+ws)/d, we get this:
3 = (s + 40)/d
Now we have our two equations. We can use algebra to solve for s, the wind speed in still air.
4d = s-40 , so d = (s-40)/4 using eq 1. Also, 3d = s+40, so d = (s+40)/3 using eq. 2
Now we have 2 equations solving for d. Since distance stays the same, we can set these equal to each other to solve for the wind speed in still air:
(s-40)/4 = (s+40)/3
3(s-40) = 4(s+40)
3s - 120 = 4s + 160
3s - 280 = 4s
Therefore, s = 280 km/h (ignore -, since only asking for speed, not direction)
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