Original Problem: Two Planes leave simultaneously from Chicago's O'Hare Airport, one flying due north and the other due east. The northbound plane is flying 50 miles faster than the eastbound plane. After 3 hours, the planes are 2440 miles apart. Find the speed of each.

So what I know is
East planes speed is = s
and north bound plane is = s + 50mph
they are forming a right triangle with the distance between them
so distance of east plane is = x
and north is = y
so y(squared) + x(squared) = 2440 miles
3 hours travel time
... I know the answer I don't care about that but I need an idea of how from the this information I can get started. Been working on this for about a hour and have come up with nothing:confused:

50 miles faster? Or further? Mile per hour? Did you check your bags or carry on?
Are you starting something? I'm not sure what your question really is...

50 miles faster? Or futher? Mile per hour? Did you check your bags or carry on?
Are you starting something? I'm not sure what your question really is....
How is it even possible to solve this question? I haven't the faintest idea where to start on it. I'm not looking for the solution I want an idea of what I should use to answer this problem.

You need to calculate where each target is at for a specific time of travel. You've given all the data necessary to create a plot: time variation vs resulting distance. Try visualizing on graph paper with a common legend for both targets.
Clue: think right angle tiangle.
You can actually start from the given range of separation after the known time span.

You need to calculate where each target is at for a specific time of travel. You've given all the data necassary to create a plot: time variation vs resulting distance. Try visualizing on graph paper with a common legend for both targets.
Clue: think right angle tiangle.
You can actually start from the given range of seperation after the known time span.
now I am confused even more, I am not sure how I am going to create a plot or I'm not sure what you mean by plot, even with your clue the pythagorean theorem (guessing that is what your hinting at) I didn't see how it would work, tried to use it many times. So from what your saying I can figure it out with time and the hypotenuse of the triangle, I am seriously not seeing it, this is just so different from a regular D=RT problem or is the formula Distance equals rate times time even used here?

Ok I figured out what I was doing wrong I was not making the connection between the legs of the triangle and the distance formula, but I figured it out thanks so much CaptainRich, what your saying makes sense now.

Hey CellistJames, I really hate burst your bubble ,BUT one thing you both forgot about, IF the aircraft flying North is flying into a headwind, his indicated airspeed willnot be the true airspeed of the aircraft::: Now to confuse you more, The one flying East Will not be true indicated airspeed either since he will be flying against a crosswind, and having to fly the nose of the aircraft towards the NORTH in order to stay on corse , Hopely he has NO tailwind from the west.. Check this out with another pilot .OR a math Prof. at a school. ;'': Good Luck, P.S. I!M not trying to be a smart alec, Just giving you some more to think about, OK

Hey CellistJames :, DID you ever get that math problem worked out?? THE ONE ABOUT THE TWO AIRPLANES// f?? F.B.E.

Hey CellistJames :, DID you ever get that math problem worked out???? THE ONE ABOUT THE TWO AIRPLANES// f??. F.B.E.
Yes I think I already said that in this post thread

This is a mixture of d=rt and Pythagoras,

The plane heading north has a rate of r.

Plane heading East is flying 50 mph slower than the one heading north, r-50

Since d=rt, they fly a distance of 3r and 3(r-50), respectively.

At which time they are 2440 miles apart.

\sqrt{(3r)^{2}+(3r-150)^{2}}=2440

Solve for r.