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

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.
-----------------
Use d = rt
r = airspeed, w = wind speed
2900 = (r-w)*300 (time in minutes)
2900 = (r - w/2)*290
-------------
29 = 3r - 3w
290 = 29r - 29w/2 --> 580 = 58r - 29w
----------
3r - 3w = 29 Eqn 1
58r-29w = 580 Eqn 2
-------------
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)
---------------------
29 = 3r - 3w
3w = 3r - 29 = 31 - 29
w = 2/3 miles/minute
w = 40 mph (headwind speed)

Answer by Theo(13342) (Show Source): 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:

(P-W) * 5 = 2900 (equation 1) becomes:
(620-40) * 5 = 2900 which becomes:
580 * 5 = 2900

Divide both sides of this equation by 5 to get:

580 = 2900/5 = 580 which is true so answer is confirmed for equation 1.

(P-W/2)*4.8333333 = 2900 (equation 2) becomes:
(620-20)*4.8333333 = 2900 which becomes:
600*4.8333333 = 2900

Divide both sides of this equation by 4.8333333 to get:

600 = 2900 / 4.8333333 = 600 which is true so answer is confirmed for equation 2.

Answer to the question is:

Airplane speed is 620 miles per hour.
Headwind speed is 40 miles per hour.

RELATED QUESTIONS
This is not for homework; however, it is one in a homework text for the University of... (answered by Alan3354,Theo)
Against a head wind, Jeff computes his flight time for a trip of 2900 miles at 5 hours.... (answered by ankor@dixie-net.com)
Against a head wind, Jeff computes his flight time for a trip of 2900 miles at 5 hours.... (answered by ankor@dixie-net.com)
Christie pilots her plane for 300 mi against a headwind in 2 hrs. The flight would take 1 (answered by Alan3354,Boreal)
An airplane took 2.5 hours to fly 625 miles with the wind. It took 4 hours and 10... (answered by piglet162431)
An airplane took 3 hr longer to fly 1200 mi than it took for a flight of 480 mi. If the... (answered by josgarithmetic)
An Airplane flew for 4 hours against headwind of 40 km/h. On the return flight the same (answered by stanbon,josmiceli)
An airplane flies 500 mi/h in still air. Flying with the jet stream, the plane travels... (answered by stanbon)
An airplane flies 900 miles against a headwind of 25 miles per hour. The plane took 15... (answered by ewatrrr)