Suppose that a wind is blowing in the direction S45°E at a speed of 50 km/h. 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.)

I got that the vectors for the wind and the pane were <50cos(45),-50sin(45)> and <150cos(60),150sin(60)> respectively. So then I added them together in order to find the true course and ground speed. But it seems this is not correct at N40.6E and 145.3km/h.

Vr=50[315oCCW] + 150[30oCCW] = Resultant

velocity.

X = 50*Cos315 + 150*Cos30 = 165.3 km.
Y = 50*sin3i5 + 150*sin30 = 39.64 km.

Tan A = Y/X = 39.64/165.3 = 0.23981.
A = 13.5o.

Vr=X/Cos A=165.3/Cos13.5 = 170km[13.5o]
CCW. = 170km[N76.5oE].

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

First, let's convert the given directions into vector form. The wind is blowing in the direction S45°E, which can be represented as a vector <-50cos(45°), 50sin(45°)> km/h.

The plane is steering in the direction N60°E, which can be represented as a vector <150cos(60°), 150sin(60°)> km/h.

To find the resultant vector of the velocities, we need to add these two vectors together:

<-50cos(45°), 50sin(45°)> + <150cos(60°), 150sin(60°)>

Calculating the values:

<-50cos(45°), 50sin(45°)> = <-35.4, 35.4> km/h

<150cos(60°), 150sin(60°)> = <75, 129.9> km/h

Adding the vectors:

<-35.4, 35.4> + <75, 129.9> = <39.6, 165.3> km/h

The resultant vector is <39.6, 165.3> km/h.

Now, to find the true course and ground speed of the plane, we will use the following trigonometric relationships:

True Course = arctan(165.3/39.6) ≈ N76.2°E

Ground Speed = magnitude of the resultant vector
Ground Speed = √(39.6^2 + 165.3^2) ≈ 169.4 km/h

Therefore, the true course of the plane is approximately N76.2°E, and the ground speed is approximately 169.4 km/h.

To find the true course and ground speed of the plane, you correctly decomposed the velocities of the wind and the plane into their vector components. However, it seems like there might be a mistake in calculating the components of the wind and plane velocities.

Let's go through the calculations step by step:

1. Wind velocity vector:
The wind is blowing in the direction S45°E (45 degrees east of south) at a speed of 50 km/h. To find the components of the wind velocity vector, we can break it down into its north-south and east-west components using trigonometry.

The north-south component, Vns, is given by:
Vns = wind speed * sin(wind angle) = 50 km/h * sin(45°) = 50 km/h * √(2)/2 = 35.4 km/h (rounded to one decimal place)

The east-west component, Vew, is given by:
Vew = wind speed * cos(wind angle) = 50 km/h * cos(45°) = 50 km/h * √(2)/2 = 35.4 km/h (rounded to one decimal place)

So the wind velocity vector is <Vew, -Vns> = <35.4, -35.4> km/h.

2. Plane velocity vector:
The plane is moving in the direction N60°E (60 degrees east of north) at an airspeed of 150 km/h. Similar to the wind velocity, let's calculate the components of the plane velocity vector.

The north-south component, Vnp, is given by:
Vnp = airspeed * cos(plane angle) = 150 km/h * cos(60°) = 150 km/h * 1/2 = 75 km/h

The east-west component, Vep, is given by:
Vep = airspeed * sin(plane angle) = 150 km/h * sin(60°) = 150 km/h * √(3)/2 = 129.9 km/h (rounded to one decimal place)

So the plane velocity vector is <Vep, Vnp> = <129.9, 75> km/h.

3. True course and ground speed:
To find the resultant vector, which represents the true course and ground speed of the plane, we add the wind velocity vector and the plane velocity vector.

Resultant vector = Wind velocity vector + Plane velocity vector
<35.4, -35.4> + <129.9, 75> = <165.3, 39.6> km/h.

The magnitude of the resultant vector represents the ground speed of the plane:
Ground speed = √(165.3^2 + 39.6^2) = 169.2 km/h (rounded to one decimal place).

Note: The true course is the direction of the resultant vector. To find it, you can use trigonometry:
True Course = arctan(Vep / Vnp)
True Course = arctan(129.9 / 75) ≈ 59.9° (rounded to one decimal place).

Therefore, the true course is approximately N59.9°E, and the ground speed is approximately 169.2 km/h.