It is a calm summer day in southeast Iowa at the Ottumwa air traffic control rad

Question

It is a calm summer day in southeast Iowa at the Ottumwa air traffic control radar installation - except there are some small, locally intense thunderstorms passing through the general area. Only two planes are in the vicinity of the station: American Flight 1003 is traveling from Minneapolis to New Orleans is approaching from the north-northwest, and United Flight 336 is traveling from Los Angeles to New York is approaching from west-southwest. Both are on the path that will take them directly over the radar tower. There is plenty of time for the controllers to adjust the flight paths to insure a safe separation of the aircraft.

Suddenly lightning strikes a power substation five miles away, knocking out the power to the ATC installation. There is, of course, a gasoline powered auxiliary generator, but it fails to start. In desperation, a mechanic rushes outside and kicks the generator; it sputters to life. As the radar screen flickers on, the controllers find that both flights are at 33,000 feet. The American flight is 32 nautical miles (horizontally) from the tower and is approaching it on a heading of 171 degrees at a rate of 405 knots. The United flight is 44 nautical miles from the tower and is approaching it on a heading of 81 degrees at a rate of 465 knots.

a. At the instant of this observation, how fast is the distance between the planes decreasing?

b. How close will the planes come to each other?

c. Will they violate the FAA's minimum separation requirement of 5 nautical miles?

d. How many minutes do the controllers have before the time of closest approach?

e. Should the controllers run away from the tower as fast as possible?

Explanation / Answer

Note tha 81 and 171 degrees are 90 degrees apart, so we can simply consider them as coming on the y and x axes.

Let American be on the y axis and United be on the x-axis

A(t) = 32 - 405t

U(t) = 44 - 465t

Then, as they are on the axes, the distances apart are f(t) = sqrt(A^2+U^2) =

sqrt((32-405t)^2+(44-465t)^2)

a) df/dt = (2(-405)(32-405t)+2(-465)(44-465t))/(2(sqrt((32-405t)^2+(44-465t)^2)))

At 0, this is

(2(-405)(32)-2(465)(44))/(2sqrt((32)^2+(44)^2)) =

-614.271816759376

It is decreasing at 614.271816759376 knots

Expand the numerator of df/dt and set it equal to 0

2(-405)(32-405t)+2(-465)(44-465t) = 0

2(-405)(32) + 2(-465)44) +2(405)^2t+2(465)^2t = 0

2(405)(32) + 2(465)44) = (2(405)^2+2(465)^2)t

t = ((405)(32) + (465)44)/((405)^2+(465)^2) = 0.0878895463510848

f(0.0878895463510848) = sqrt((32-405t)^2+(44-465t)^2) =

sqrt((32-405*0.0878895463510848)^2+(44-465*0.0878895463510848)^2) =

4.76774170302309

c)Yes, they violate the 5 nautical mile rule, as 4.76774170302309 < 5

d) 0.0878895463510848 hours = 0.0878895463510848*60minutes =

5.27337278106509 minutes

e) Even though they will be within 5 nautical miles, 4.76774170302309 nautical miles is still far away. There is no reason to run from the tower.

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