You are working as an assistant to an air-traffic controller at the local airport, from which small airplanes take off and land.

Question

3 months ago

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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 18.0 km, and 25.0° south of west. The second aircraft is at altitude 1,100 m, horizontal distance 20.0 km, and 20.0° south of west.
1. Your supervisor is concerned that the two aircraft are too close together and asks for a separation distance (in km) for the two airplanes. (Place the x-axis west, the y-axis south, and the z-axis vertical.)

Physics

1 answer:

serg [3.5K] · Answer 1 · 2026-03-11T18:43:10+03:00

Answer:

d = 2021.6 km

Explanation:

This distance problem can be solved using vector analysis; it's best to find each plane's position components before applying the Pythagorean theorem to calculate the separation between them.

For Airplane 1:

Height y₁ = 800m

Angle θ = 25°

cos 25 = x / r

sin 25 = z / r

x₁ = r cos 20

z₁ = r sin 25

x₁ = 18 103 cos 25 = 16,314 103 m = 16314 m

z₁ = 18 103 sin 25 = 7,607 103 m = 7607 m

For Plane 2:

Height y₂ = 1100 m

Angle θ = 20°

x₂ = 20 103 cos 25 = 18.126 103 m = 18126 m

z₂ = 20 103 sin 25 = 8.452 103 m = 8452 m

To determine the distance between the planes using the Pythagorean theorem:

d² = (x₂-x₁)² + (y₂-y₁)² + (z₂-z₁)²2

Now, we perform the calculations:

d² = (18126-16314)² + (1100-800)² + (8452-7607)²

d² = 3,283 106 + 9 104 + 7,140 105

d² = (328.3 + 9 + 71.40) 10⁴

d = √(408.7 10⁴)

d = 20,216 10² m

d = 2021.6 km