KMA-628

O.K., I am posting a math question where I disagree with the answer. Here is the question:

An aircraft with a constant speed in still air travels 4200 km with a constant tailwind in 3 hours. With the same wind now against the aircraft, it takes 4 hours to make the return trip. What was the speed of the plane, in km/hr?
A. 875
B. 125
C. 1225
D. 175
E. None of these

I am being told that the tailwind/headwind needs to be factored into the equation to make it a problem with two variables (a systems problem). I disagree since the tailwind is the exact opposite of the headwind it should cancel itself out.

To me (I am not a math guru) the answer should be 1050km/hr which would make the answer to the above problem "E".

My answer is wrong. Any suggestions?

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EK69

i get 1225
is that correct?
(also, your reasoning is correct, however, it wont cancel out...or it doesnt in my calculation....lol)

KMA-628

yes, that is the correct answer.

I have a problem with the wording.

The plane is not really traveling at 1225km/hr it is travelling at 1225km/hr minus headwind. Since headwind equals 175km/hr the "actual" speed of the plane is 1050km/hr.

I think it should've said, "What is the speed of the plane before the headwind/tailwind is accounted for?"

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EK69

yes the wind (head or tail) is 175km/h

what you have to take into account is the direction of flight.

this is how i did it:

going: distance = 4200km ; Net speed = plane speed + tail wind (it is pushing the plane forward) ; time = 3 hrs

coming: distance = 4200km ; Net speed = plane speed - head wind (it is pushing the plane backwards) ; time = 4hrs

from this stems 2 equations

4200 = 3 (plane+tail)
4200 = 4 (plane -wind)

solving these equations you get plane speed to be 1225km/h and wind speed to be 175km/h

so that means the net speed going was 1400km/h and the net speed coming was 1050km/h, however, the plane is still going at 1225km/h.

Dawson70

Ek is correct based on the numerical question. The equations work out.
However, if the question is to be inturpited, the answer is E

This turns into a calculas equation based on the resistance of the wind.
Wind=SpeedxT(thrust)/ (.7234/100km=R) Resistance/ Weight
Formulate D/M x R (T/1.3533) (.7234x100km)= KPH

I came up with 945KM....not an option above.

Sorry if I confused any one. I have a way of complicating things

But in theory, you have to many missing variables in the question.
What is the weight of the aircraft, including fuel.
What is the thrust, etc etc etc

I only mention this, because while I was in college (advanced calc), many questions presented themselves very familiar to this one. However, the Professor would expect his students to perform the numerical equations and they would be wrong. He would want the students to question the question to make it feasable with the missing variables.
Arggggg.....never mind

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EK69

i get what you are saying.
i mean if you think about it, the aircrafts mass would be pulled towards earth, therefore making it sort of a parabolic/curvilinear motion question. you'd have to also factor in normal air drag (even if its still)

this oculd be over complicated to high level calculus/dynamics, however, for the sake of making it damn near impossible to solve, i guess we assume a "point particle" as the plane (as in, having in infintesimally small mass, which can then be neglected, etc)


typos are cuz im drunk and just got home from clubbin lol

Peetster

That's contradictory. Still air doesn't move.

KMA-628

Peester,

That is what I thought.

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EK69

lol yeah i thought that was weird too, but then figured they meant more in terms of no air drag caused by wind going in any direction other than parallel to the planes flight

Dawson70

Still air only exsists in space. Then it is not really air. Of course no tail wind. If this is a question for a class. I would put my money on answer E. Too many unknowns. Math professors either love me or hate me......

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rsl12

not a very clear question. it might be better to ask: what does the speedometer read inside the plane?

speaking of which, how *do* speedometers work inside planes? is it just an airflow meter?

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Phage

"An aircraft with a constant speed in still air travels 4200 km with a constant tailwind in 3 hours."

I interpreted this as: "There is an aircraft that moves at a constant airspeed when there is no wind. In this instance, it travels 4200 km in 3 hours while it has a tailwind."

A tailwind will never be cancelled out by an equal headwind on the return trip because the duration of the flights is different. A flight with a tailwind will be shorter than with the headwind, and this changes the amount of time the wind has to effect the aircraft.

rsl12

by the way, you are correct when you say that the tailwind and headwind cancel each other out **when you consider the round trip flight as a whole**. In other words, it takes 7 hours to get from here to there, and the head and tail winds cancel each other out. therefore, you travel 8400 kilometers in 7 hours, and you divide 8400 by 7 to get the velocity of the plane.

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KMA-628

"C" is the correct answer.

I agree with pretty much everybody here in that the wording on this question probably creates more questions.

Yes, it came from an Algebra test. When I first saw this one it seemed easy: Divide the distance by the time and you get the speed. For my logic process, how can the plane be traveling at 1225km/hr and only go 4200km in four hours?

I compare it to driving.

If I am going 60mph in my car with cruise control on, for four hours straight (without stopping) I would travel 240 miles. Drag, wind speed, tailwind/headwind, drafting behind a semi, etc. makes no difference. I still travel 240 miles in four hours.

The only thing that changes is the efficiency of my trip, i.e. mpg.

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kutulu

Personally, I have a problem with teachers using things they don't understand in order to make a real world application possible. In order to assume that the constant wind has the same effect traveling both ways you have to assume that wind coming from the front has the same effect as wind coming from behind. The only way that could happen is if the plane was perfectly symmectical.

Comparing it to a car, cars are designed to minimize drag from the front of the vehicle. The engineers are not concerned with how a tailwind effects the vehicle because it does not reduce efficiency. If you put a car in a wind tunnel you'd get completely different flow patterns when you change the wind direction. The drag coefficients are totally different.

I just hate it when people misuse concepts like this. To me, its like saying the ice cools the water rather than being more correct by saying that the warm water melts the ice.

rsl12

kutulu, this is incorrect. a plane will be push by two horizontal forces (yes you philosophy majors, there are more than 2 forces, i don't need to hear about it): the force exerted by the engines, and the force of drag. If the plane is going at a constant speed, that means that the two forces cancel each other out. assuming that the weight of the plane and the force of engines remain the same in both directions, we only need to look at drag. drag is a function of the speed that the plane is travelling relative to the air about it. whether or not the wind is a tailwind or headwind is irrelevant, because the plane will reach the same velocity relative to the wind, no matter whether it's coming from the front or behind. (winds coming from the side are a different matter.) in other words:

Fe+Fd = 0

where Fe = vector force of the engine
Fd = vector force of drag, which is a function of Vp, vector velocity of the plane relative to the air around it (the turbulence of which we assume is constant). If the engines are turned off and the plane miraculously still floats, and the (laminar) air is going 500 mph, the plane will (eventually) also go 500 mph, so that the equation balances. Basically, the plane will go as fast as necessary to balance this equation (ie, as fast as necessary to achieve the Vp (velocity vector) value that satisfies the equation. The equation is unsolvable if the wind is not exactly facing towards or away from the plane, or if the plane is rising or sinking. (Then you need to take into account the rudders or whatever else planes use to steer themselves). But other than those issues, the equation is independent of the wind speed relative to the ground.

Basically, what I'm saying is that the question asked is perfectly fine as it is, even though it's worded poorly.

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rsl12

or to say it in an easier way, you can draw the control volume for analyzing the plane's movement without including the ground. therefore, the ground, and anything measured relative to it, make absolutely no difference in determining the plane's velocity relative to the air.

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kutulu

Drag forces are a hell of a lot more complicated than that. They are highly dependant on geometry and the velocity of the object, relative to the fluid.

rsl12

yes yes kutulu, i know that but all that stuff is considered constant in this problem, no matter which way you are going.

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kutulu

Constant or not, you can't break the equation down into:

dist = time1 * (plane speed + wind speed)
dist = time2 * (plane speed - wind speed)

Although the drag force will be stronger on the return trip, it won't "slow the plane down" by 175 km/hr and on the first flight you won't get a 175 km/hr boost from the wind. Unless of course we have wind speeds approaching the speed of sound...

EK69

lol im guessing this is meant to be an introductory algebra class type question, not a dynamics class question where you have to take into account every force that exists.

they like to idealize the situation as much as possible, so u can actually calculate it out.

kutulu

Idealizing a situation is fine, but when the question is going to require a ridiculous and flat out wrong assumption it is a bad question. There are many other ways in which you could present the same type of problem where you set up a system of two equations. The professor could have used the classic two trains traveling in opposite directions or maybe a problem involving a see-saw or something like that where you could neglect the wieght of the boards and friction.

rsl12

kutulu, it CAN be broken down into those two equations, with very basic assumptions about the way the plane is flying. Going both ways, the plane will be travelling exactly the same speed relative to the air around it. Drag and all other relevant forces will be functions of quantities that will be exactly the same in both directions. The drag force will NOT be stronger on the return trip--it will be exactly the same. there is no reason for the drag to be greater, since the air molecules will be passing by the airplane at the same velocity as when the plane was going to wherever it was going.

Again, consider this scenario--a boat is travelling downstream a very very deep and wide river (deep and wide because I don't want viscocity and distance to ground to be an issue, as it isn't with an airplane other than to determine turbulance and drag issues. the river can be a bit turbulent--that's fine). If the boat has no engine on it, then the steady state solution is that it goes exactly the same speed as the river water, if the water is not turbulent, and some slightly slower speed if the water is turbulent (and i remind you that turbulence is a scalar quantity). if it were going any faster, drag would slow it down. if it were going any slower, drag would speed it up.

If the boat had an engine that could read a speed of 10 mph, then the boat could go downstream at a speed of 10 mph + the river current speed. If it were going upstream, it would be 10 - the current speed. same reasoning.

EDIT: actualy, with turbulence, the boat I was wrong. the boat will go either slightly slower *or* slightly faster, as it gets joslted a bit in all directions, so you have a random walk effect.

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Dawson70

speaking of which, how *do* speedometers work inside planes? is it just an airflow meter? [/B][/QUOTE]

They are called pedo tubes. The take the outside pressure and calculate the pressure forced into the tube.
With the new GPS systems in the aircraft now, the speed is calculated between the signal beacons.
The out side pressure of the aircraft is regulated by the Barometric Pressure. Example: 29.92 would be mostly a good sunny day. High pressure. A low front could register at 26.05.
The altitude of the aircraft is depended on this pressure. Once an aircraft reaches approx 18,000ft, regardless of the pressure below, the altimeter goes back 29.92.

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rsl12

thanks for the info Dawson. Makes sense--that's how you measure water flow through a channel as well!

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rlbond86

Oh come on guys it's an algebra question.

(a) Plane's speed + wind = 4200 / 3 = 1400
(b) Plane's speed - wind = 4200 / 4 = 1050

subtract (b) from (a)

2 * wind = 1400 - 1050 = 350
wind = 175

back to (a)

plane's speed + wind = plane's speed + 175 = 4200 / 3 = 1400
plane's speed = 1400 - 175 = 1225
C


Yes we can say that the drag force would be proportional to the velocity, that planes don't actually go straight, but in great circles, that the drag would decrease as altitude increases, and get a not-so-nice differential equation. But the guy did want an answer.

Oh, and don't forget relativity!

Tachion

The problem states it takes 4 hours for the return trip. Now is that from looking at the pilots watch or from the clock at the beginning airport and another at the end of the trip.
The plane could have been going through a time zone.


But all that is to fool you.
The real speed of the plane was zero as it was stated it completed the trip.

xepherys

if 2 x wind = 250, wouldn't wind then = 125?

rlbond86

silly me.
i got confused.
see my post for corrections