An airplane is flying on a compass heading​ (bearing) of 330° at 320 mph. A wind is blowing with the bearing 300° at 40 mph.

Question

An airplane is flying on a compass heading​ (bearing) of 330° at 320 mph. A wind is blowing with the bearing 300° at 40 mph.

​(a) Find the component form of the velocity of the airplane. ​

(b) Find the actual ground speed and direction of the plane.

Answer 1

(a) The component form of the velocity of the airplane. It can be found by considering the horizontal and vertical components but separately.

The horizontal component is the speed of the airplane multiplied by the cosine of the angle between its heading and the x-axis. That means, it would be 320 mph * cos(30°) because the bearing of 330° is 30° counterclockwise from the x-axis.

The vertical component is the speed of the airplane multiplied by the sine of the angle between its heading and the x-axis. It would be 320 mph * sin(30°).

(b) To find the actual ground speed and direction of the plane, we need to add the effects of the wind. We can use vector addition to determine the resultant velocity.

The horizontal component of the wind is the wind speed multiplied by the cosine of the angle between its bearing and the x-axis. In this case, it would be 40 mph * cos(60°) because the bearing of 300° is 60° counterclockwise from the x-axis.

The vertical component of the wind is the wind speed multiplied by the sine of the angle between its bearing and the x-axis. In this case, it would be 40 mph * sin(60°).

Next, we add the horizontal components of the airplane's velocity and the wind's velocity, and we add the vertical components of the airplane's velocity and the wind's velocity separately.

The actual ground speed is the magnitude of the resultant velocity vector, which can be found using the Pythagorean theorem.

The direction of the plane's movement is the angle that the resultant velocity vector makes with the positive x-axis, which can be found using trigonometry.

I hope this helps! If you have any further questions, feel free to ask.

Answer 2

The airplane's velocity: V_airplane = 320 mph

The direction of the airplane (converted from compass bearing to standard position): θ_airplane = 360° - 330° = 30°

The x and y components of the airplane's velocity (V_ax and V_ay) can be calculated as follows:

V_ax = V_airplane * cos(θ_airplane) = 320 mph * cos(30°) = 320 mph * (√3 / 2) = 277.13 mph

V_ay = V_airplane * sin(θ_airplane) = 320 mph * sin(30°) = 320 mph * (1 / 2) = 160 mph

Similarly, the wind's velocity: V_wind = 40 mph

The direction of the wind (converted from compass bearing to standard position): θ_wind = 360° - 300° = 60°

The x and y components of the wind's velocity (V_wx and V_wy) can be calculated as follows:

V_wx = V_wind * cos(θ_wind) = 40 mph * cos(60°) = 40 mph * (1 / 2) = 20 mph

V_wy = V_wind * sin(θ_wind) = 40 mph * sin(60°) = 40 mph * (√3 / 2) = 34.64 mph

(b) The actual ground speed and direction of the plane are determined by adding the airplane's velocity components and the wind's velocity components.

The total velocity in the x direction (V_tx) is the sum of the x-components of the airplane's velocity and the wind's velocity:

V_tx = V_ax + V_wx = 277.13 mph + 20 mph = 297.13 mph

The total velocity in the y direction (V_ty) is the sum of the y-components of the airplane's velocity and the wind's velocity:

V_ty = V_ay + V_wy = 160 mph + 34.64 mph = 194.64 mph

The actual ground speed (V_t) can be found using the Pythagorean theorem:

V_t = sqrt((V_tx)^2 + (V_ty)^2) = sqrt((297.13 mph)^2 + (194.64 mph)^2) = 350.94 mph

The actual direction (θ_t) can be found using the arctangent function, bearing in mind that the result from the arctangent function might need adjustment based on which quadrant the result is in:

θ_t = atan(V_ty / V_tx) = atan(194.64 mph / 297.13 mph) = 33.39°

However, since we are interested in the compass bearing, we need to convert this back from standard position:

Compass bearing = 360° - θ_t = 360° - 33.39° = 326.61°

So, the actual ground speed is approximately 350.94 mph and the compass bearing is 326.61°.