My first answer would be 129 miles, on the assumption that “flies” means "travels through the air (in response to the various collective forces in effect). But since this is an algebra question, I’d grudgingly do the Pythagorean thing.
Ideally, you could express your answer as a range, from 129 to 138.351.
What I think is confusing you is that the wind doesn’t affect the plane’s movement on the north-south axis. The wind is only causing the plane to drift along the east-west axis, and that’s why the total distance increases.
See the diagram in post #7 for an illustration.
Did the plane take off from a treadmill?
Yes and thats why when the illustration was provided I understood it better. I didn’t like how the question was phrased
No but I heard there were snakes on it 
I suppose I am nitpicking but shouldn’t the question have some disclaimer that they are going to ignore the surface area/wind resistence of the plane and how that will affect its path?
I hated quesions like this because they always left something out.
The other poorly worded part of the problem is a “50 mph wind from the west”.
Just because the wind is blowing at 50mph doesn’t mean it will push a plane along at 50 mph.
Better wording would be “a wind from the west pushes the plane westward at 50mph”.
Yabbut, either you interpret the problem strictly as it’s stated, in which case the correct answer is “129 miles,” or you assume that it has an airspeed of 129 due North, with a 50 mph Westerly wind, in which case you treat it exactly as the Pythagorean model.
[uqote=ultrafilter]What I think is confusing you is that the wind doesn’t affect the plane’s movement on the north-south axis. The wind is only causing the plane to drift along the east-west axis, and that’s why the total distance increases.
[/quote]
If the wind did affect the north-south axis, the total distance would change as well, and even more.
I dunno. There are pretty much two ways to interpret the 120 mph north of the OP: it’s either ground speed, in which case the rest of the problem is classic red herring–the ground speed is the ground speed, period. But if it is the air speed, then it’s relative to the air, and a 50 mph movement of air would move the plane 50 mph in addition to the 120 mph.
If you interpret “flies north” as meaning that the plane’s axis points north then it will cover 130 miles over the ground in one hour. If you interpret it as meaning that the ground track is north then it will cover 109 miles in one hour. The latter is because in order to fly a north ground track the plane must head slightly to the west and thus into a slight headwind.
Oh yes it does. The plane is immersed in a fluid and is carried along by that fluid. As long as the plane’s velocity in the direction of the wind is less than the velocity of the wind there will be a force that accelerates it.
[QUOTE=Greenback]
I suppose I am nitpicking but shouldn’t the question have some disclaimer that they are going to ignore the surface area/wind resistence of the plane and how that will affect its path?QUOTE]
That is all irrelevant. As David said, the aircraft is fully immersed in the air and any movement of the air over the ground will take the plane with it. A 50mph wind from west will move the plane east at 50mph, no more, no less.
But would the force necessarily be 50 mph’s worth? Imagine a plane with a sail, like a sailboat. Wouldn’t the effect of the wind vary depending on which way the sail was turned?
I’d start a thread called “Snakes on a Plane on a Treadmill” but I know better 
Let’s take this incrementally. When the plane first takes off with a 50 mph cross wind there is the force of a 50 mph wind accelerating it toward downwind. After some increment of time the plane is now going 10 mph downwind and there is a net 40 mph force accelerating it. After another increment of time the plane is goind 20 mph downwind and there is a net 30 mph wind accelerating it. And so on.
Eventually the plane will be going 49 mph downwind and there will be a 1 mph wind accelerating it. And the process will continue until the plane is going 50 mph downwind and there is a 0 mph wind at which time, and not before, the downwind acceleration will be zero.
PS - It’s nothing at all like the sail of a sailboat. The sailboat has its hull and especially that big keel or centerboard in the water and that provides considerable resistance so the boat doesn’t go downwind so fast. Just go out in a centerboard boat and pull up the centerboard and watch what happens.
An airplane has no such wind-countering force.
A sailboat is a bad analogy. A sailboat has part of its structure exposed to the air, and part of its structure in the water. The resultant track of the sailboat is the outcome of the"fight" between the force exerted by the air (wind) and the force exerted by the sea.
An aircraft is only in the air, there is nothing to “fight” the wind and so the wind’s effect is full.
If you are sitting on a flat-bed train carriage, you go exactly the same speed as the train in the direction the train is moving. If it was a wide carriage and you started walking directly across the carriage, you are now moving exactly the same speed as the train, in the direction the train is moving, you are also moving at your walking pace across the carriage. The total distance and direction travelled is the reultant of those two vectors.
Like the person walking on the carriage, the aircraft is entirely contained in the air. If the air is moving west at 50mph across the ground, then the aircraft’s movement due to the wind will be exactly 50 mph west across the ground. It is also moving at 120mph to the north and so the total distance travelled is the resultant of those two vectors, or the hypotenuse of the triangle they form.
And on preview I see that David has covered it, not to worry.
And, just for the sake of completeness, any process in which the output is proportional to the difference between a fixed input, 50 mph in this case, and the present state of the output, *i.e.*the various velocities of the plane, results in an exponential curve.
Plane velocity = wind velocity(1 - e[sup]-kt[/sup]). k is a constant depending on the pounds of force on the plane for a given velocity difference and the mass of the plane. The exponential term will eventually reach zero and the plane velocity will equal the wind velocity. The constant k is called the time constant (or its reciprocal) and it will be quite short. If you watch a plane, even a 747, take off in a brisk crosswind you will see it rapidly pick up the crosswind velocity.
What a horribly worded question. A dumb person will not know how to do the problem, an average student will use the pythagorean formula and get the “right” answer, a smart student will agonize over the wording and try and figure out what is meant and come up with one of three possible answers, two of which will be marked wrong. Too bad you can’t flunk the text book company.
Which is a perfectly understandable assumption, for those of us used to ground-based transport with wheels on. I distinctly remember arguing this point from the wrong side when I was at school (but WHY can’t the wind just be blowing PAST the plane?) The subtleties of physics (things stop because of friction and/or drag, energy isn’t created or destroyed but you still use it up and get billed for it, etc.) aren’t that intuitive. Even when you think you’re on top of them they can still bite you, try reading the “plane on a treadmill” threads! (Where I was wrong again…)