A plane is headed due west with an air speed of 212 km/h. It it driven from its course by a wind from due north blowing at 23.6 km/h. Find ground speed of the plane and the actual direction of travel.
The ground speed is the vector sum of 212 km/h west and 23.6 km/h south.
Ground speed = sqrt[(212)^2 + 23.6)^2] = 213.3 km/h
Direction = tan^-1(23.6/212) S of W
That is 6.35 degrees S of W, or a bearing 263.65 degrees
To find the ground speed of the plane and its actual direction of travel, we can use vector addition. The ground speed is the magnitude of the resultant vector, and the direction is the angle that the resultant vector makes with the due west direction.
To solve this problem, we can break down the velocities of the plane and the wind into their horizontal (x) and vertical (y) components.
1. Start by drawing a coordinate system with the positive x-axis pointing east and the positive y-axis pointing north.
- Draw a vector pointing due west with a length of 212 km/h to represent the air speed of the plane.
- Draw a vector pointing due north with a length of 23.6 km/h to represent the wind speed.
2. Now, break down the plane's airspeed vector into its x and y components:
- The x-component of the plane's airspeed is -212 km/h (negative because it is pointing west).
- The y-component of the plane's airspeed is 0 km/h (since it is not pointing north or south).
3. Similarly, break down the wind vector into its x and y components:
- The x-component of the wind is 0 km/h (since it is not pointing east or west).
- The y-component of the wind is 23.6 km/h (positive because it is pointing north).
4. To find the resultant vector, add the x-components and y-components separately:
- The x-component of the resultant vector is -212 km/h + 0 km/h = -212 km/h.
- The y-component of the resultant vector is 0 km/h + 23.6 km/h = 23.6 km/h.
5. Use the Pythagorean theorem to find the magnitude of the resultant vector, which represents the ground speed of the plane:
- Ground speed = √((-212 km/h)^2 + (23.6 km/h)^2)
- Ground speed = √(44,944 km^2/h^2 + 556.96 km^2/h^2)
- Ground speed = √45,500.96 km^2/h^2
- Ground speed ≈ 213.53 km/h (rounded to two decimal places).
6. To find the direction of travel, use trigonometry. The angle, θ, can be found using the tangent function:
- tan(θ) = (opposite / adjacent) = (23.6 km/h / 212 km/h)
- θ ≈ tan^(-1)(0.111)
- θ ≈ 6.36 degrees (rounded to two decimal places).
7. Since the plane is headed due west, the actual direction of travel will be 180 degrees minus the angle, θ:
- Actual direction of travel ≈ 180 degrees - 6.36 degrees
- Actual direction of travel ≈ 173.64 degrees (rounded to two decimal places).
Therefore, the ground speed of the plane is approximately 213.53 km/h, and its actual direction of travel is approximately 173.64 degrees.