SOLUTION: A helicopter flew 15 miles against a 25 mph headwind. Then it flew back with the wind at its tail. The round trip lasted 27 minutes. Find the helicopter's speed in still air.
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Question 937771: A helicopter flew 15 miles against a 25 mph headwind. Then it flew back with the wind at its tail. The round trip lasted 27 minutes. Find the helicopter's speed in still air.
You can put this solution on YOUR website! x = time with headwind
y = time with tailwind
z = helicopter speed in still air
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s = d/t
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speed in headwind:
z - 25 = 15/x
z = 15/x + 25
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speed in tailwind:
z + 25 = 15/y
z = 15/y - 25
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equate z's:
15/x + 25 = 15/y - 25
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total time:
x + y = 27/60
y = 27/60 - x
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15/x + 25 = 15/y - 25
15/x - 15/y = -50
y(15/x - 15/y = -50)
15y/x - 15 = -50y
15(27/60 - x)/x - 15 = -50(27/60 - x)
(15*27/60)/x - 15 - 15 = -50*27/60 + 50x
(15*27/60)/x = -50*27/60 + 30 + 50x
6.75/x = 7.5 + 50x
50x - 6.75/x + 7.5 = 0
50xx + 7.5x - 6.75 = 0
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the above quadratic equation is in standard form, with a=50, b=7.5 and c=-6.75
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to solve the quadratic equation, by using the quadratic formula, copy and paste this:
50 7.5 -6.75
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the quadratic has two real roots at:
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x = 0.3
x = -0.45
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the negative root doesn't fit the problem statement, so use the positive root:
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x = 0.3 hr
y = 27/60 - x
y = 27/60 - 0.3
y = 0.15 hr
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answer:
z = 15/x + 25
z = 15/0.3 + 25
z = helicopter speed in still air = 75 mph
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You can put this solution on YOUR website!
A helicopter flew 15 miles against a 25 mph headwind. Then it flew back with the wind at its tail. The round trip lasted 27 minutes. Find the helicopter's speed in still air.
Let speed in still air be S
Then time taken to fly against the wind =
Time taken to fly with the wind =
Since total trip time was 27 minutes, or , or hr, then:
15(20)(S + 25) + 15(20)(S – 25) = 9(S – 25)(S + 25) ----- Multiplying by LCD, 20(S – 25)(S + 25)
(S – 75)(3S + 25) = 0
S, or speed in still air = mph OR S = (ignore)
