Question 813579: a plane travels at a speed of 210mph in still air. Flying with a tailwind the plane is clocked over a distance of 675.flying against a headwind it take one more hour time to complete the return trip.what was the wind velocity?
Answer by TimothyLamb(4379)   (Show Source):
You can put this solution on YOUR website! p = 210
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t = d / s
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downwind flight:
a = 675/(p + w) ... X
b = 675/(p - w)
b = a + 1
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upwind flight:
b = 675/(p - w)
a + 1 = 675/(p - w)
a = 675/(p - w) - 1 ... Y
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X = Y
a = 675/(p + w) = a = 675/(p - w) - 1
675/(p + w) = 675/(p - w) - 1
1 = 675/(p - w) - 675/(p + w)
1 = 675(p + w)/(p - w)(p + w) - 675(p - w)/((p - w)(p + w))
1 = ( 675(p + w) - 675(p - w) )/((p - w)(p + w))
1 = ( 675p + 675w - 675p + 675w )/((p - w)(p + w))
1 = ( 675w + 675w )/((p - w)(p + w))
1 = 1350w/((p - w)(p + w))
recall that p = 210
1 = 1350w/((210 - w)(210 + w))
1 = 1350w/(44100 - ww)
44100 - ww = 1350w
ww + 1350w - 44100 = 0
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the above quadratic equation is in standard form, with a=1, b=1350, and c=-44100
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to solve the quadratic equation, plug this:
1 1350 -44100
into this: https://sooeet.com/math/quadratic-equation-solver.php
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Answer 1:
the quadratic has two real roots, but one is negative, so use the positive root:
w = 31.9 mph
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you can also find the times for downwind and upwind flights by solving the above equations for a and b:
a = 2.79 hours
b = 3.79 hours
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