We had a problem in our text book which we did in class.
The details were
If a plane is traveling due North at 120 MPH and there is a wind from the West at 50 MPH how far will the plane travel in one hour?
The instructor drew a grid on the board and using the therom that is
2 2 2
C=A +B
(sorry I don’t know hows to do superscripts)
He drew a triangle and determined the plane went 150 miles. I can not understand how a wind that is going across the plane can make it go futher than it’s normal rate. I can understand if the wind was pushing the plane because it was comming from behind but I don’t see that happening in this sceneario
It seems to be a very poorly defined question. It doesn’t clarify whether the 120mph is ground speed or air speed, and it doesn’t clarify whether “how far” refers to distance along the ground or relative to the air. And more importantly, we don’t know if “traveling due north” means its actual path (relative to the ground) is due north, or whether the plane is pointed north but due to the wind but drifts towards the east.
Nevertheless, the point of the question is that the plane’s path along the ground is the vector sum of the plane’s motion relative to air and the motion of the air (wind). It’s like crossing a river in a boat. Say the river is 1 mile wide and flowing at 5 mph, and you row your boat at 5mph. If you point your boat at right angle to the river, you reach the opposite shore after 1/5 hour of rowing. You’re also 1 mile downstream of where you started. So the actual distance you travelled is 1 mile across the river and 1 mile downstream, which adds up to sqrt(1[sup]2[/sup] + 1[sup]2[/sup]) = sqrt(2) miles. On the other hand, if you turn your boat against the current to make sure you don’t end up downstream, you have to turn your boat 45 degrees, as if you are heading for a point 1 mile upstream of where you are. Relative to the water, you need to row sqrt(2) miles, so it takes sqrt(2)/5 hours to get to the other shore. You end up 1 mile from where you started (across the river but not downstream).
I’m not sure how much more simply I can explain it; your teacher drew a pretty clear picture. The wind moves the plane fifty miles. The plane moves the plane 120 miles. The hypotenuse or those two distances is 150 miles.
Consider a more extreme example, which might make it easier to understand. Suppose you have a long movator - a moving sidewalk, like you see in an airport, okay? Suppose it’s 200 feet long and four feet wide, and it takes one minute to ride to the end.
Now suppose I have a little turtle. He’s very slow. I’ve trained him that whenever he’s put on a movator, he wants to walk ACROSS the movator, e.g. from side to side. It takes my slow little turtle one minute to walk the four feet.
So I put him down on the movator over on one side of it. He obediently lurches across the movator. At the same time, the movator moves him all the way down to the end, 200 feet, while he walks the four feet across.
So the TURTLE’S speed was four feet per minute. Meanwhile, the movator, moving across him (e.g. at a right angle to his own direction of movement) moved him 200 feet. So he moved 200.0399 feet (the hypotenuse of 200 and 4.)
So there you have a force, the movator, moving at a right angle to the turtle, moving him further than his own speed. Just as you have a force, the wind, moving at a right5 angle to the plane, moving it more than its own speed.
Well, if the plane is “traveling due north at 120mph”, then it’s darn well moving at 120mph in a northerly direction, so it will travel 120 miles due north in one hour.
If the plane is pointed due north with an airspeed of 120mph, then the 50mph westerly cross wind will make the plane travel northeast at a speed greater than 120mph, in relation to the ground below. What your teacher did was break up the actual velocity of the plane into its component vectors, 120mph north and 50mph east, then combine them with the formula to find the total velocity. I’m getting a velocity of 130 though, not 150.
I’m sure someone will be along to correct my mistake, or verify that I was right.
It’s an algebra problem, guys. It’s not about airspeed vs. groundspeed or anything else that actually matters in a real-life situation. It’s basically just looking for the student knowing how to apply the Pythagorean theorem.
Cheessteak is right: as with all too many book questions, the people writing it did not actually stop to think what they were saying, and simply assumed you would apply whatever you just learned in class blindly.
Are you sure the answer was 150? I got a different answer.
Anyway, the answer I got still had the plane going further than 120 miles.
As your teacher showed, it all comes down to vectors, and the vectors can be shown as a triangle with 50 along one side, 120 along another at right angles, with the resultant distance travelled being the third side. The Pythagorean Theorem describes how to work out the longer side as I’m sure your teacher showed you.
I’m not sure how to explain to you how to get your head around the plane travelling further with a wind at right angles. All I can say is that if it travels 120 miles north and gets blown 50 miles east then it definitely ends up more than 120 miles from its start point.
For the wind to blow such that the distance travelled by the plane is the same, it must blow slightly towards the aircraft.
Rickjay is on the right track explaining it I think.
Try this. The plane is moving 50mph and the wind is at 120mph at right angles. In an hour, the wind moves the plane 120 miles and the engine moves the plane 50 miles, result? 130 miles travelled, a whole lot further than the 50 mph would’ve taken it without the wind.
I agree entirely with Cheesesteak. I have no idea where the 150 figure came from; I can only assume it’s a typo. I’d be worried about any math teacher that came up with that as the correct figure.
I think the other two posters do a good job of explaining away the OP’s confusion, so I’ll leave it at that.
If the teacher got an answer of 150, ask him what he was on. Of course, the intended answer was 130, but the question is brutally unclear.
In lowest terms what you have is a 5,12,13 right triangle. I am reminded of a friend of mine who acted as a helper when some local construction people were putting up his (literally) log cabin. The spread of the roof was 24 feet, the peak was centered 5 feet above the top of the walls and he told the builders they would need 14 foot logs (there was a one foot overhang). They were astounded that anyone could know that, since they had been building all their lives and would simply have taken 18 foot logs (that’s 12 + 5 + 1) and then trimmed them all. It was not merely that they didn’t know the formula, they didn’t know that a formula was even possible.
But the problem that (I think) the book is intending to ask the kids to solve is exactly the classic groundspeed vs. airspeed one. That is what it’s about.
But being the smartass I am, I would have also pointed out that if the plane is “travelling due North at 120 mph” then after one hour it’s gone 120 miles.
I think the question was poorly constructed. I think I’d phrase it that a plane could go 120 mph north with no wind. How far would it go in one hour with a 50 mph wind from the east? Or better yet, point the plane north on a huge treadmill moving to the east.
I get the impression that the OP wasn’t quoting the problem from his textbook word-for-word, just giving the gist. It may be that the problem was actually worded a bit more carefully, so that Cheesesteak’s objection wouldn’t apply. But I agree that, as presented in the OP, it’s poorly worded and ambiguous, since “traveling due north” could well mean that the pilot was compensating for the wind in such a way as to make the plane move in a northerly direction.
I’d hope the design of the plane wouldn’t allow the wind to blow it west at 50mph with perfect efficiency. Unless it’s a sailplane, heh.
For the OP, superscripts and subscripts are easy with the {sup} and {sub} tabs:
A{sup}2{/sup} + B{sup}2{/sup} = C{sup}2{/sup}
Replace all curly brackets with squares ones and the screen will show:
A[sup]2[/sup] + B[sup]2[/sup] = C[sup]2[/sup]
Make sure to preview, though. It’s easy to miss something and have your formula appear as:
From Introductory Algebra A Just-in-Time Approach Edition 3 by Alice Kaseberg page 8
Question #2
A plane flies due north at 129 miles per hour (mph). There is a 50-mph cross wind from the west. How many miles will the plane travel in 1 hour?
Thats the entire question and I thank everyone for the answers they have given me. My problem was I was picturing the wind from the west just blowing over the plane and not helping it’s advancement like a tail wind would and not hindering it as a head wind would.
