Expert: Scotto - 4/10/2010

Question
You have two flights A(Minneapolis to New Orleans) and B(Los Angeles to New York). The flight paths indicate that they are going to intersect over a city. Both flights are at 33,000 ft, flight A was 32 nautical miles from the city approaching at a heading of 171degrees at a rate of 405 knots. At the same time flight B is 44 nautical miles from the city approaching at a heading of 81degrees at a rate of 465 knots.

They want to know how fast the planes are approaching each other, and if they will violate the minimum separation requirement of 5 nautical miles rule. If possible how close the planes actually get to each other, and the time they are closest.Would Air Control have time to take appropriate action? Might a slight altitude change for one of the flights help prevent any five mile rule violation?

Thank you for any information, I'm having trouble starting the problem even after drawing a diagram of it.

Answer
Take the city as the origin on a graph.
Take north-south as the y axis and east-west as the x-axis.

Put A on the y axis, 32 nautical miles away, approaching at 405 knots.
Thus, the position of A is 32 - 405t.

Note that the difference in degrees is 90, so the other plane is on the x axis.
Its position is 44 - 465t.

To draw take a sheet of graph paper.  Put A as 32 in the y direction with an arrow pointing
towards the origin.  Put B at 44 on the x axis with an arrow point towards the origin.

The distance between them is then √((32-405t)² + √(44-465t)²).

To find the minimum, take the derivative of this and set it equal to 0.
Note that the derivative is [2(32-405t)405 - 2(44-465t)(-465)]/√((32-405t)² + √(44-465t)²).

The denominator can be ignored, since denominators never make fractions 0
and it is never 0 in this case.  Take 2(32-405t)405 - 2(44-465t)(-465),
multiply it out, set it to 0, and solve for t.

At this value of t will be the minimum distance, so put this t back in the distance equation to find out how far they were apart.