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 200 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.)

Trouble-free way to get resultant:

resolve into x- and y-components and add.
Wind is FROM N45W, which means that its direction is S45E.
Take East = x, and North = y,
the wind vector is then
<40sin(45),40cos(45)>

The velocity vector of the plane is
<200sin(60),200cos(60)>

The combined vector (resultant) is therefore
<40sin(45)+200sin(60), 40cos(45)+200cos(60)>
Find the magnitude and direction of the resultant in the usual way.

41.6

To find the true course and ground speed of the plane, we need to find the resultant vector of the wind velocity and the airspeed of the plane.

1. Convert the wind direction angle N45°W to the standard form:

N45°W = 45° west of north = 360° - 45° = 315°

2. Convert the plane direction angle N60°E to the standard form:

N60°E = 60° east of north = 360° - 60° = 300°

3. Break down the wind velocity and airspeed vectors into their components.

The wind velocity vector can be broken down into its north and west components:

Wind velocity (Vw) = 40 km/h

Vw_north = Vw * sin(315°) = 40 km/h * sin(315°)
= 40 km/h * (-0.7071) [approximating sin(315°) as -0.7071]
= - 28.3 km/h (south)

Vw_west = Vw * cos(315°) = 40 km/h * cos(315°)
= 40 km/h * 0.7071 [approximating cos(315°) as 0.7071]
= 28.3 km/h (west)

The airspeed vector can be broken down into its north and east components:

Airspeed (Va) = 200 km/h

Va_north = Va * sin(300°) = 200 km/h * sin(300°)
= 200 km/h * 0.5 [approximating sin(300°) as 0.5]
= 100 km/h (north)

Va_east = Va * cos(300°) = 200 km/h * cos(300°)
= 200 km/h * 0.866 [approximating cos(300°) as 0.866]
= 173.2 km/h (east)

4. Add the corresponding components of the wind velocity and airspeed vectors to get the resultant vector.

Resultant north component = Vw_north + Va_north
= -28.3 km/h + 100 km/h
= 71.7 km/h (north)

Resultant east component = Va_east - Vw_west
= 173.2 km/h - 28.3 km/h
= 144.9 km/h (east)

5. Use the resultant vector components to calculate the true course and ground speed.

True Course (TC) = arctan(Resultant east component / Resultant north component)
= arctan(144.9 km/h / 71.7 km/h)
= arctan(2.018)

TC ≈ 63.9°

Ground Speed (GS) = √(Resultant north component^2 + Resultant east component^2)
= √(71.7 km/h)^2 + (144.9 km/h)^2)
= √(5152.89 km^2/h^2 + 21005.01 km^2/h^2)
= √(26157.9 km^2/h^2)
= 161.6 km/h

So, the true course of the plane is approximately N63.9°E and the ground speed is approximately 161.6 km/h.

To find the true course and ground speed of the plane, we need to determine the resultant of the velocity vectors of the plane and the wind.

Step 1: Convert the given wind direction and speed into vector form.
The wind is blowing from N45°W at a speed of 40 km/h. The vector representing the wind can be obtained by decomposing it into its northward (N) and westward (W) components.

The northward component (N-W component) = wind speed * sin(wind direction)
= 40 km/h * sin(45°)
≈ 40 km/h * 0.7071
≈ 28.28 km/h

The westward component (W-W component) = wind speed * cos(wind direction)
= 40 km/h * cos(45°)
≈ 40 km/h * 0.7071
≈ 28.28 km/h

Thus, the vector representing the wind is (28.28 km/h N, 28.28 km/h W).

Step 2: Convert the given airspeed and direction of the plane into vector form.
The plane is steered in the direction N60°E at an airspeed of 200 km/h. We can also decompose this vector into its northward (N) and eastward (E) components.

The northward component (N component) = airspeed * sin(plane direction)
= 200 km/h * sin(60°)
= 200 km/h * 0.866
≈ 173.2 km/h

The eastward component (E component) = airspeed * cos(plane direction)
= 200 km/h * cos(60°)
= 200 km/h * 0.5
= 100 km/h

Thus, the vector representing the plane's velocity is (173.2 km/h N, 100 km/h E).

Step 3: Determine the resultant of the wind and plane vectors.
To find the resultant vector, we simply add the corresponding components of the plane and wind vectors together.

Resultant vector = (Plane N-component + Wind N-component, Plane E-component + Wind E-component)
= (173.2 km/h N + 28.28 km/h N, 100 km/h E + 28.28 km/h W)
= (201.48 km/h N, 71.72 km/h E)

Step 4: Find the magnitude of the resultant vector.
The magnitude of the resultant vector represents the ground speed of the plane.

Ground speed = √((N-component)^2 + (E-component)^2)
= √((201.48 km/h)^2 + (71.72 km/h)^2)
≈ √(40598.6304 km^2/h^2 + 5142.6784 km^2/h^2)
≈ √(45741.3088 km^2/h^2)
≈ 213.9 km/h

Therefore, the true course of the plane is N60°E and the ground speed is approximately 213.9 km/h.