An air-traffic controller observes two aircraft on his radar screen. The first is at altitude 750 m, horizontal distance 20.0 km, and 25.0° south of west. The second aircraft is at altitude 1050 m, horizontal distance 17.0 km, and 17.0° west of south.
(a) Write the displacement vector FROM the first plane TO the second plane, letting i represent east, j north, and k up.
i?
j?
k?
(b) How far apart are the two planes?
To find the displacement vector from the first plane to the second plane, we need to break it down into its horizontal (x and y) components as well as its vertical (z) component.
First, let's find the horizontal components. The given information tells us that the first plane is 20.0 km away at a 25.0° angle south of west, while the second plane is 17.0 km away at a 17.0° angle west of south.
To find the x component, we need to find the horizontal distance in the east-west direction. We can use trigonometry to break down the distances into their respective x and y components.
For the first plane:
x₁ = 20.0 km * cos(25.0°) (east is positive)
x₁ = 18.478 km (east)
For the second plane:
x₂ = 17.0 km * sin(17.0°) (west is negative)
x₂ = -4.625 km (west)
To find the y component, we need to find the horizontal distance in the north-south direction.
For the first plane:
y₁ = 20.0 km * sin(25.0°) (north is positive)
y₁ = 8.350 km (north)
For the second plane:
y₂ = 17.0 km * cos(17.0°) (south is negative)
y₂ = -15.978 km (south)
Now, let's find the vertical component (z). The given information tells us the altitudes of the two planes.
For the first plane:
z₁ = 750 m
For the second plane:
z₂ = 1050 m
Now we have all the components needed to write the displacement vector.
(a) The displacement vector from the first plane to the second plane can be written as:
i = x₂ - x₁ = (-4.625 km) - (18.478 km) = -23.103 km (west)
j = y₂ - y₁ = (-15.978 km) - (8.350 km) = -24.328 km (south)
k = z₂ - z₁ = (1050 m) - (750 m) = 300 m (up)
Therefore, the displacement vector from the first plane to the second plane is -23.103 km (west) i, -24.328 km (south) j, +300 m (up) k.
(b) To find the distance between the two planes, we can use the Pythagorean theorem. The distance is the magnitude of the displacement vector.
Distance = sqrt(i² + j² + k²)
Distance = sqrt((-23.103 km)² + (-24.328 km)² + (300 m)²)
Distance ≈ 36.016 km
Therefore, the two planes are approximately 36.016 km apart.
To find the displacement vector from the first plane to the second plane, we can break it down into its i, j, and k components.
(a) The i-component represents the east-west direction, the j-component represents the north-south direction, and the k-component represents the up-down direction.
From the given information, we know that the first plane is 20.0 km west and 25.0° south of west, while the second plane is 17.0 km south and 17.0° west of south.
To find the i-component, we need to consider the horizontal distance in the west-east direction. Given that the first plane is to the west and the second plane is to the southwest, the i-component will be negative.
i = -(20.0 km) * cos(25.0°)
To find the j-component, we need to consider the horizontal distance in the north-south direction. Given that the first plane is south of west and the second plane is to the south, the j-component will also be negative.
j = -(17.0 km) * cos(17.0°)
The k-component represents the difference in altitude between the two planes. Given that the first plane is at 750 m and the second plane is at 1050 m, the k-component can be calculated as:
k = 1050 m - 750 m
Now we can find the displacement vector from the first plane to the second plane:
Displacement vector = i + j + k
(b) To find the distance between the two planes, we can calculate the magnitude of the displacement vector.
Distance = √(i^2 + j^2 + k^2)