Distance between two aircraft

Hi,

Hoping someone can help me with this problem.

Aircraft A and B are traveling perpendicularly toward each other. Aircraft A's velocity is 320 mph, and aircraft B's velocity is 160 mph. At 1:00 PM, aircraft A is 800 miles from the destination, and aircraft B is 410 miles from the same destination.

A.) Obtain the expression for the distance D between the aircraft as a function of time. B.) Use the Roots function to compute the time when the aircraft are first within 30 miles of each other.

Of course, d=r*t. I can compute easily the time the two aircraft meet at the destination, which is 2.52 hours. However, the 30 miles apart bit is throwing me. I know I need to set up a polynomial by setting the two d=r*t equations equal to each other, and then operate on that using the Roots function, but for the life of me I'm stumped as to how to put it all together.

Any help is certainly much appreciated !

Chris

Answer 1

[Roger Stafford]

Christopher, you have made a wrong assumption. These planes will not arrive at the destination point at the same time. By the slightly ambiguous phrase "traveling perpendicularly toward each other" you can assume that the airplanes' two routes toward their common destination are at right angles. What you need to do is compute the distance between the planes as a function of the time past 1:00 PM. Think back on your geometry days in high school and see what formula you learned there that would allow you to make this computation. It is very straightforward. Once you have done that, you can formulate an equation by setting this equal to 30. It is fairly easy to so manipulate this equation that it will be in a form that matlab's 'roots' function can solve, (although as it happens you also presumably learned in high school another method of solving such an equation that doesn't even require 'roots'.)

[Christopher]

Thank you Roger, much appreciated. I see the error I made in thinking that the two planes arrive at the same time. It's obvious that they don't.

With regards to geometry, I'm assuming you are referring to using the Pythagorean Theorem. I thought of that, as it is indeed a right triangle with the hypotenuse equal to 30 miles. I wasn't sure how to solve sides a and b as a function of time/position.

I'm thinking that A1's equation is 800-320t, and A2's equation is 410-160t. If I plug these values into the Pythagorean Theorem for "a" and "b" respectively, Matlab should be able to solve for the roots (time 't' when the two aircraft are 30 miles apart).

Am I on the right track ?

Chris

[Roger Stafford]

You are indeed on the right track, Christopher. As the hypotenuse length, you will get the square root of stuff in t which is to be set equal to 30. Note that the matlab 'roots' function demands that you have the coefficients of a polynomial in a certain special arrangement, and this square root isn't in the right form for that at all. You now have to figure out how to manipulate this annoying square root so as to get some kind of polynomial equation in t and find all its coefficients. Then you can use 'roots'. Or alternatively when you look at the polynomial you arrive at, you should be able to recognize it as a kind of algebra problem you solved in school without even using matlab.

It may surprise you that there will be two solutions, one of which would not apply to the given problem. But if the planes didn't land and continued on in straight lines past the airport with the same speeds, there would come a time they were again 30 miles apart, so the second solution is not as crazy as it looks. Good luck!

[Christopher]

Got it. I computed the two roots to be 2.432 and 2.593.

The first root is the time after 1PM to the first time the aircraft are 30 miles apart, which after converting to time format gives 2 hours, 26 minutes. This puts the first time at 3:26PM.

The second 30 mile distance occurs at t=2.593, or 2 hours, 36 minutes, which puts the second event at 3:36PM.

Feels like its right.

Thanks again for all your help Roger !