Question 592619: an airplane flying into a headwind travels the 1800-mile flying distance between two cities in 3 hours and 36 minutes. On the return flight, the distance is traveled in 3 hours. Find the air speed of the plane and the speed of the wind, assuming that both remain constant.
Answer by lwsshak3(11628)   (Show Source):
You can put this solution on YOUR website! an airplane flying into a headwind travels the 1800-mile flying distance between two cities in 3 hours and 36 minutes. On the return flight, the distance is traveled in 3 hours. Find the air speed of the plane and the speed of the wind, assuming that both remain constant.
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let x=speed of plane
let c=speed of wind
x-c =net speed into wind
x+c=net speed with wind
speed*travel time=distance
3 hrs and 36 min=3.6 hrs
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(x-c)*3.6=1800
(x+c)*3=1800
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3.6x-3.6c=1800
3.0x+3.0c=1800
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multiply first equation by 3 and second equation by 3.6
10.8x-10.8c=5400
10.8x+10.8c=6480
add equations to eliminate c
21.6x=11880
x=11880/21=550 mph
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3c=1800-3x=1800-1650=150
c=150/3=50 mph
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ans:
speed of plane=550 mph
speed of wind=50 mph
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