Question

The magnitude of a velocity vector is called speed. Suppose that a wind is blowing from the direction N45°W at a speed of 50 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 150 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.)

Answer 1

Answer:

True course = 49.4° = N40.6°E

Ground speed = 145.32 km/h

Step-by-step explanation:

Velocity of the wind = (vₓî + vᵧj) km/h

Magnitude = 50 km/h

Direction = N45°W

Velocity of the wind = 50 [(-sin 45°)î + (cos 45°)j] = 50 (-0.7071î + 0.7071j) = (-35.36î + 35.36j) km/h

Velocity of the plane in still air = (vₓî + vᵧj) km/h

Magnitude = 150 km/h

Direction = N60°E

Velocity of the plane in still air = 150 [(sin 60°)î + (cos 60°)j] = 150 (0.866î + 0.5j) = (129.9î + 75j) km/h

Resultant velocity = vector sum of the two velocities

R = (-35.36î + 35.36j) + (129.9î + 75j)

R = (94.54î + 110.36j) km/h

True course = direction of the resultant

True course = tan⁻¹ (110.36/94.54)

True course = 49.4° = N40.6°E

Ground speed = magnitude of the resultant velocity

Ground speed = √(94.54² + 110.36²) = 145.32 km/h