Question

42. You are working as an assistant to an air-traffic controller at the local airport, from which small airplanes take off and land. Your job is to make sure that airplanes are not closer to each other than a minimum safe separation distance of 2.00 km. You observe two small aircraft on your radar screen, out over the ocean surface. The first is at altitude 800 m above the surface, horizontal distance 19.2 km, and 25.08 south of west. The second aircraft is at altitude 1 100 m, horizontal distance 17.6 km, and 20.08 south of west. Your supervisor is concerned that the two aircraft are too close together and asks for a separation distance for the two airplanes. (Place the x axis west, the y axis south, and the z axis vertical.)

Answer 1

Separation distance between two points P (x1, y1, z1) and Q (x2, y2, z2) is given by:

PQ = sqrt ((x2 - x1)^2 + (y2 - y1)^2 + (z2 - z1)^2)

Now Assuming the x axis west, the y axis south, and the z axis vertical

Now given that Position of 1st plane is:

P = 800 m Above the surface and distance 19.2 km, at 25.08 degree south of west

x1 = 19.2*cos 25.08 deg = 17.39 km in west

y1 = 19.2*sin 25.08 deg = 8.14 km in south

z1 = 800 m = 0.8 km

P = (17.39, 8.14, 0.8) km

Q = 1100 m Above the surface and distance 17.6 km, at 20.08 degree south of west

x2 = 17.6*cos 20.08 deg = 16.53 km in west

y2 = 17.6*sin 20.08 deg = 8.14 km in south

z2 = 1100 m = 1.1 km

Q = (16.53, 6.04, 1.1) km

Now distance between both planes will be:

PQ = sqrt ((17.39 - 16.53)^2 + (8.14 - 6.04)^2 + (0.8 - 1.1)^2)

PQ = 2.29 km

Since aircrafts are at a distance of 2.29 km, which is greater than 2 km, So they are at a safe distance.

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