David W. answered • 05/12/15
Tutor
4.7
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Experienced Prof
I was an army helicopter pilot in Vietnam and I wouldn’t ride with any pilot who could not solve this problem because they would probably get lost when they flew too far or not far enough!
First, realize that the aircraft speed is relative to the surrounding air (and it is often measured in nautical miles per hour), and maybe not land miles per hour. So, in my helicopter, if I wanted to hover over one spot on land and there was a 20 mile/hour wind, I would have to fly at 20 miles/hour (in relation to the air around me) in order to maintain my position.
Helicopters and airplanes usually take off and land facing into the wind because that means slower ground speed is required before takeoff and you will reach slower ground speed sooner for landing (be sure to think about this!).
Your problem asks you to find the speed of the plane in still air. That means it is constant and the units of speed are miles/hour (or mph). Also, in order to fly the two distances given, THE SAME LENGTH OF TIME is required (this is very important because we need this information to make an equation (=) so we can find the aircraft speed).
Let x be the speed at which our plane files relative to still air. Then, in the “same time,” our plane flies:
405 miles = (x +20) miles/hour * t hours and 315 miles = (x – 20)miles/hour * t hours
We don’t need t yet, and we could have initially set up the ratio as:
405 / (x+20) = 315 / (x-20)
Now, do cross multiplication”
405 * (x -20) = 315 * (x + 20)
405x – 8100 = 315x + 6300
90x = 14400
x = 160 miles per hour
Checking (super important because everybody makes errors and Vietnam is no place to get lost -- I did once:
Flying downwind, t = 405 / (160+20) = 405 / 180 = 2.25 hours
Flying into the wind, t = 315 / (160-20) = 315 / 140 = 2.25 hours
Yep, that’s the same amount of time!
First, realize that the aircraft speed is relative to the surrounding air (and it is often measured in nautical miles per hour), and maybe not land miles per hour. So, in my helicopter, if I wanted to hover over one spot on land and there was a 20 mile/hour wind, I would have to fly at 20 miles/hour (in relation to the air around me) in order to maintain my position.
Helicopters and airplanes usually take off and land facing into the wind because that means slower ground speed is required before takeoff and you will reach slower ground speed sooner for landing (be sure to think about this!).
Your problem asks you to find the speed of the plane in still air. That means it is constant and the units of speed are miles/hour (or mph). Also, in order to fly the two distances given, THE SAME LENGTH OF TIME is required (this is very important because we need this information to make an equation (=) so we can find the aircraft speed).
Let x be the speed at which our plane files relative to still air. Then, in the “same time,” our plane flies:
405 miles = (x +20) miles/hour * t hours and 315 miles = (x – 20)miles/hour * t hours
We don’t need t yet, and we could have initially set up the ratio as:
405 / (x+20) = 315 / (x-20)
Now, do cross multiplication”
405 * (x -20) = 315 * (x + 20)
405x – 8100 = 315x + 6300
90x = 14400
x = 160 miles per hour
Checking (super important because everybody makes errors and Vietnam is no place to get lost -- I did once:
Flying downwind, t = 405 / (160+20) = 405 / 180 = 2.25 hours
Flying into the wind, t = 315 / (160-20) = 315 / 140 = 2.25 hours
Yep, that’s the same amount of time!
