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 subtract the position vector of the first plane from the position vector of the second plane.
(a) Let's first find the i-component of the displacement vector.
The horizontal distance from the first plane to the second plane is the difference between their respective horizontal distances.
i-component = (17.0 km - 20.0 km) = -3.0 km (negative sign indicates westward direction)
The j-component is found in a similar way:
The horizontal distance from the first plane to the second plane is the difference between their respective horizontal distances.
j-component = (25.0 km - 17.0 km) = 8.0 km (positive sign indicates northward direction)
Since the displacement vector is from the first plane to the second plane, it means they are moving in the opposite directions.
The k-component is the altitude difference between the two planes.
k-component = (1050 m - 750 m) = 300 m (positive sign indicates upward direction)
Therefore, the displacement vector from the first plane to the second plane is:
i = -3.0 km
j = 8.0 km
k = 300 m
(b) To find the distance between the two planes, we can use the Pythagorean theorem:
distance = √(i^2 + j^2 + k^2)
distance = √((-3.0 km)^2 + (8.0 km)^2 + (300 m)^2)
distance = √(9.0 km^2 + 64.0 km^2 + 90,000 m^2)
distance = √(73.0 km^2 + 90,000 m^2)
Converting 73.0 km to meters, we get:
distance = √(73,000 m^2 + 90,000 m^2)
distance = √(163,000 m^2)
distance = 403.12 m
Therefore, the two planes are approximately 403.12 meters apart.
To find the displacement vector from the first plane to the second plane, we need to determine the components of the vector in the east (i), north (j), and up (k) directions.
(a) Let's start by determining the components of the displacement vector:
For the first plane:
Altitude: 750 m
Horizontal distance: 20.0 km (convert to meters by multiplying by 1000)
Direction: 25.0° south of west
The horizontal and vertical components of the displacement vector for the first plane can be found using trigonometry.
Horizontal component (i):
The horizontal distance is given by 20.0 km * cos(25.0°). Convert it to meters by multiplying by 1000.
i = 20.0 km * cos(25.0°) * 1000
Vertical component (k):
The altitude is directly the vertical component.
k = 750 m
Now let's determine the components for the second plane:
For the second plane:
Altitude: 1050 m
Horizontal distance: 17.0 km (convert to meters by multiplying by 1000)
Direction: 17.0° west of south
Horizontal component (i):
The horizontal distance is given by 17.0 km * sin(17.0°). Convert it to meters by multiplying by 1000.
i = 17.0 km * sin(17.0°) * 1000
Vertical component (k):
The altitude is directly the vertical component.
k = 1050 m
Finally, we can find the displacement vector from the first plane to the second plane by taking the differences between the corresponding components:
Displacement vector (from first to second plane):
i = (i_2 - i_1)
j = 0 (since there is no north-south displacement)
k = (k_2 - k_1)
(b) To determine how far apart the two planes are, we can use the magnitude of the displacement vector. The magnitude is calculated using the Pythagorean theorem:
Magnitude = sqrt(i^2 + j^2 + k^2)
Substitute the values of i, j, and k that we calculated to find the magnitude. This will give us the distance between the two planes.