The magnitude of a velocity vector is called speed. Suppose that a wind is blowing from the direction N45°W at a speed of 40 km/h. (This means that the direction from which the wind blows is 45° west of the northerly direction.) A pilot is steering a plane in the direction N60°E at an airspeed (speed in still air) of 100 km/h. The true course, or track, of the plane is the direction of the resultant of the velocity vectors of the plane and the wind. The ground speed of the plane is the magnitude of the resultant. Find the true course and the ground speed of the plane. (Round your answers to one decimal place.)
We can think of this problem as the sum of two vectors. Let vector A represent the wind direction, and vector B represents the plane direction.
Let's first find their components. The wind speed A is 40 km/h and its direction is N45°W. Therefore we can write its components as:
A_x = -40 cos 45° (west is negative)
A_y = 40 sin 45° (north is positive)
A_x = -40 * (1/√2) = -20√2
A_y = 40 * (1/√2) = 20√2
The airspeed of the plane B is 100 km/h and its direction is N60°E. Therefore we can write its components as:
B_x = 100 cos 60°
B_y = 100 sin 60°
B_x = 100 * (1/2) = 50
B_y = 100 * (√3/2) = 50√3
Now we can find the components of the resultant vector by simply adding the corresponding components of vectors A and B:
R_x = A_x + B_x = -20√2 + 50
R_y = A_y + B_y = 20√2 + 50√3
Now let's find the magnitude (ground speed) and direction (true course) of the resultant vector R.
Ground speed (magnitude of R) = √(R_x^2 + R_y^2) = √((-20√2 + 50)^2 + (20√2 + 50√3)^2) ≈ 104.5 km/h
True course (direction of R) = tan^(-1)(R_y / R_x) = tan^(-1)((20√2 + 50√3) / (-20√2 + 50)) ≈ 82.9°
This angle is measured from the east direction, so the true course is N(90 - 82.9)°E = N7.1°E.
Thus, the ground speed is approximately 104.5 km/h, and the true course is N7.1°E.