Question 527804: a plane flies 500 miles with the wind and 300 miles against the wind in the same length of time if the speed of the wind is 28mph, what is the speed of the plane in still air
Answer by lmeeks54(111)   (Show Source):
You can put this solution on YOUR website! Let v = the velocity of the airplane in still air in miles per hour
Let t = the number of hours the airplane flies
Let d = the distance the airplane flies in miles
28 = wind speed in miles per hour
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General rate, distance formula:
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d = v * t
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500 = (v + 28) * t
300 = (v - 28) * t
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The easiest way to solve this: make both equations equal to the same thing. The easiest way to do that is to set them = 0. For example:
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500 = (v + 28) * t can be simplified by subtracting 500 from both sides:
0 = vt + 28t - 500
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300 = (v - 28) * t can be simplified by subtracting 300 from both sides:
0 = vt -28t - 300
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Since both equations are equal to zero, they are equal to each other. Now we have:
vt + 28t - 500 = vt - 28t - 300
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We can continue to simplify:
- First subtract the vt term from both sides:
28t - 500 = -28t -300
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- Then add 28t to both sides:
56t - 500 = -300
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- Then add 500 to both sides:
56t = 200
t = 200/56
t = ~3.57 hrs
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We now take value for t and substitute it back into any of our original equations to solve for v. After that, we will check our work:
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500 = vt + 28t
500 = 3.57v + 28(3.57)
500 = 3.57v + 100
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Subtract 100 from both sides:
400 = 3.57v
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Divide both sides by 3.57:
v = 112
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This is the answer to the problem. All we need to do is to check our work.
300 = vt - 28t
300 = (112)(3.57) - 28(3.57)
300 = 400 - 100
300 = 300 checks
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cheers,
Lee
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