SOLUTION: Ben is a piolet for Air Airwas. He computes his flight time against a headwind for a trip of 2900 mi at 5 hr. The flight would take 4 hr and 50 min if the headwind were half as g
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Question 229129: Ben is a piolet for Air Airwas. He computes his flight time against a headwind for a trip of 2900 mi at 5 hr. The flight would take 4 hr and 50 min if the headwind were half as great. Find the headwind and the planes air speed. Found 2 solutions by Alan3354, Theo:Answer by Alan3354(69443) (Show Source):
You can put this solution on YOUR website! Ben is a piolet for Air Airwas. He computes his flight time against a headwind for a trip of 2900 mi at 5 hr. The flight would take 4 hr and 50 min if the headwind were half as great. Find the headwind and the planes air speed.
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Use d = rt
r = airspeed, w = wind speed
2900 = (r-w)*300 (time in minutes)
2900 = (r - w/2)*290
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29 = 3r - 3w
290 = 29r - 29w/2 --> 580 = 58r - 29w
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3r - 3w = 29 Eqn 1
58r-29w = 580 Eqn 2
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87r - 87w = 841 Eqn 1 x 29
174r- 87w = 1740 Eqn 2 x 3
------------------ Subtract
-87r = -899
r = 31/3 miles/minute
r = 620 mph ( = airspeed)
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29 = 3r - 3w
3w = 3r - 29 = 31 - 29
w = 2/3 miles/minute
w = 40 mph (headwind speed)
You can put this solution on YOUR website! Let P = speed of the plane.
Let W = Speed of a full headwind.
Let P = Speed of the plane.
Let R1 = Overall speed going against a full headwind = P-W
Let R2 = Overall speed going against half a full headwind = P-(W/2)
Let T1 = Amount of time it takes against a full headwind.
Let T2 = Ampount of time it takes against half a full headwind.
Let D = distance.
Distance is 2900 miles
With a full headwind it takes 5 hours.
With half a full headwind it takes 4 hours 50 minutes = 4.8333333 hours.
We have:
T1 = 5 hours
D = 2900
T2 = 4.8333333 hours
Rate * Time = Distance
Formula against a full headwind is:
R1 * T1 = D which becomes:
R1 * 5 = 2900
Since R1 = P-W, this formula becomes:
(P-W) * 5 = 2900 (equation 1)
Formula against half a full headwind is:
R2 * T2 = D becomes:
R2 * 4.8333333 = 2900
Since R2 = (P-W/2), this formula becomes:
(P-W/2)*4.8333333 = 2900 (equation 2)
We can solve for (P-W) to get:
(P-W) = 2900 / 5 = 580 miles per hour.
We can solve for (P-W/2) to get:
(P-W/2) = 2900 / 4.8333333 = 600 miles per hour.
Since (P-W) = 580, we can solve for P to get:
P = W + 580 (equation 3)
Since (P-W/2) = 600, we can solve for P to get:
P = W/2 + 600 (equation 4)
Since equation 3 and equation 4 both equal to P, then they both equal to each other and we get:
W + 580 = W/2 + 600
Multiply both sides of this equation by 2 to get:
2W + 1160 = W + 1200
Subtract W from both sides of this equation and subtract 1160 from both sides of this equation to get:
2W - W = 1200 - 1160
Combine like terms and simplify to get:
W = 40
Substitute 40 for W in equation 3 to get:
P = W + 580 (equation 3) becomes:
P = 40 + 580 = 620
So far we have:
P = 620
W = 40
To confirm these answers are correct, we substitute in equations 1 and 2 to get:
