Question 931415: A helicopter can fly 110 mph in still air. When flying to a town 560 miles away and returning, the helicopter has to fly with a tailwind of x mph when going to the town and with a head wind of x mph when returning. If the return flight takes 3 hours more than the flight to the town, find the wind speed. Thank you so much for whoever can help me out.
Found 2 solutions by ankor@dixie-net.com, TimothyLamb:
Answer by ankor@dixie-net.com(22740)   (Show Source):
You can put this solution on YOUR website! A helicopter can fly 110 mph in still air.
When flying to a town 560 miles away and returning, the helicopter has to fly with a tailwind of x mph when going to the town and with a head wind of x mph when returning.
If the return flight takes 3 hours more than the flight to the town, find the wind speed.
:
From the information given we know
(110+x) = ground speed with the wind
and
(110-x) = ground speed against the wind
:
Write a time equation; time = dist/speed
headwind time - tailwind time = 3 hrs
 -  = 3
solve for x
Answer by TimothyLamb(4379)  (Show Source):
You can put this solution on YOUR website! x = wind speed
t = time flying to the town
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s = d/t
t = d/s
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to town:
t = 560/(110 + x)
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returning from town:
t + 3 = 560/(110 - x)
t = 560/(110 - x) - 3
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equate times:
560/(110 + x) = 560/(110 - x) - 3
560/(110 + x) - 560/(110 - x) = -3
560(110 - x)/(110 + x)(110 - x) - 560(110 + x)/(110 + x)(110 - x) = -3
560(110 - x) - 560(110 + x) = -3(110 + x)(110 - x)
560*110 - 560*x - 560*110 - 560*x = 3xx - 3*110*110
-2*560*x = 3xx - 3*110*110
3xx + 2*560*x - 3*110*110 = 0
3xx + 1120x - 36300 = 0
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the above quadratic equation is in standard form, with a=3, b=1120 and c=-36300
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to solve the quadratic equation, by using the quadratic formula, copy and paste this:
3 1120 -36300
into this solver: https://sooeet.com/math/quadratic-equation-solver.php
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the quadratic has two real roots at:
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x = 30
x = -403.333333
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the negative root doesn't fit the problem statement, so use the positive root:
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answer:
x = wind speed = 30 mph
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