The magnitude of a velocity vector is called speed. Suppose that a wind is blowing from the direction N 45° W at a speed of 30 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 N 60° E at an airspeed (speed in still air) of 300 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.)

To find the true course and the ground speed of the plane, we need to calculate the resultant velocity vector, which is the vector sum of the plane's velocity vector and the wind's velocity vector.

### Step 1: Represent the Velocity Vectors
- **Wind's velocity vector**: The wind is blowing from N 45° W, which means it is going towards S 45° E. The angle for the vector is 45° south of east, which corresponds to 135° from the positive x-axis (East direction).
- **Plane's velocity vector**: The plane is flying towards N 60° E, so its angle from the positive x-axis is 60°.

We can represent the wind and plane velocities in their respective components.

### Step 2: Calculate Components of the Vectors

- **Wind's velocity components**:
  - \( V_{wx} = 30 \cos(135^\circ) \)
  - \( V_{wy} = 30 \sin(135^\circ) \)

- **Plane's velocity components**:
  - \( V_{px} = 300 \cos(60^\circ) \)
  - \( V_{py} = 300 \sin(60^\circ) \)

### Step 3: Add the Vectors to Find the Resultant

- Resultant velocity components:
  - \( V_{rx} = V_{px} + V_{wx} \)
  - \( V_{ry} = V_{py} + V_{wy} \)

### Step 4: Compute Ground Speed and True Course
- **Ground speed** (magnitude of the resultant vector):
  \[
  \text{Ground Speed} = \sqrt{V_{rx}^2 + V_{ry}^2}
  \]
  
- **True course** (angle of the resultant vector):
  \[
  \text{True Course} = \arctan\left(\frac{V_{ry}}{V_{rx}}\right)
  \]

Let's calculate these values.

### Step 5: Perform the Calculations

1. **Wind's velocity components**:
   - \( V_{wx} = 30 \cos(135^\circ) = 30 \times (-\frac{\sqrt{2}}{2}) = -21.21 \, \text{km/h} \)
   - \( V_{wy} = 30 \sin(135^\circ) = 30 \times \frac{\sqrt{2}}{2} = 21.21 \, \text{km/h} \)

2. **Plane's velocity components**:
   - \( V_{px} = 300 \cos(60^\circ) = 300 \times \frac{1}{2} = 150 \, \text{km/h} \)
   - \( V_{py} = 300 \sin(60^\circ) = 300 \times \frac{\sqrt{3}}{2} = 259.81 \, \text{km/h} \)

3. **Resultant velocity components**:
   - \( V_{rx} = 150 + (-21.21) = 128.79 \, \text{km/h} \)
   - \( V_{ry} = 259.81 + 21.21 = 281.02 \, \text{km/h} \)

4. **Ground speed**:
   \[
   \text{Ground Speed} = \sqrt{128.79^2 + 281.02^2} \approx \sqrt{16580.19 + 78973.64} \approx \sqrt{95553.83} \approx 309.1 \, \text{km/h}
   \]

5. **True course**:
   \[
   \text{True Course} = \arctan\left(\frac{281.02}{128.79}\right) \approx \arctan(2.18) \approx 65.9^\circ
   \]

Thus:
- **True course**: \( N 65.9^\circ E \)
- **Ground speed**: \( 309.1 \, \text{km/h} \)

### Conclusion:
- **True Course**: \( N 65.9^\circ E \)
- **Ground Speed**: \( 309.1 \, \text{km/h} \)

Would you like more details or explanations on any part of the process?

Here are five questions to deepen your understanding:
1. How do we determine the direction of the wind's vector from its description?
2. What is the significance of converting angles to standard positions from the positive x-axis?
3. Why do we need to break vectors into components for addition?
4. How does the resultant vector influence the plane's true course?
5. What would happen to the ground speed if the wind's speed increased?

**Tip:** Understanding vector components is essential for solving problems involving forces, velocity, and other directional quantities in physics.

Explore how to calculate the true course and ground speed of a plane using vector addition. This problem involves trigonometry and vector components to solve real-world kinematics problems. Suitable for students in Grades 10-12.

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