An airplane has a ground speed of 350 km/hr in the direction duewest. If there is a wind blowing northwest at 40 km/hr, calculate thetrue air speed and heading of the airplane.

To calculate the true airspeed and heading of the airplane, we can use vector addition. The problem involves two vectors: the airplane's ground speed vector and the wind speed vector.

### Step 1: Define the Vectors
1. **Ground Speed Vector (Airplane's Speed Relative to the Ground)**:
   - Magnitude: 350 km/hr
   - Direction: Due West (270°)

2. **Wind Speed Vector**:
   - Magnitude: 40 km/hr
   - Direction: Northwest (45° from the North)

### Step 2: Break Down the Vectors into Components
We break down the vectors into their north-south and east-west components.

#### Ground Speed Vector:
- Since the direction is due West, it only has a component in the west (or negative east) direction.
- **East-West Component (X-axis)**: \( V_x = -350 \) km/hr
- **North-South Component (Y-axis)**: \( V_y = 0 \) km/hr

#### Wind Speed Vector:
- Northwest is at a 45° angle from the North, so the components are equal.
- **East-West Component (X-axis)**: \( V_x = -40 \times \cos(45^\circ) = -40 \times \frac{1}{\sqrt{2}} \approx -28.28 \) km/hr
- **North-South Component (Y-axis)**: \( V_y = 40 \times \sin(45^\circ) = 40 \times \frac{1}{\sqrt{2}} \approx 28.28 \) km/hr

### Step 3: Add the Components to Find the True Airspeed Vector
Now we add the corresponding components of the ground speed vector and the wind speed vector:

#### East-West Component (X-axis):
\[ V_{x, \text{total}} = -350 + (-28.28) = -378.28 \text{ km/hr} \]

#### North-South Component (Y-axis):
\[ V_{y, \text{total}} = 0 + 28.28 = 28.28 \text{ km/hr} \]

### Step 4: Calculate the True Airspeed and Heading
1. **True Airspeed** (Magnitude of the Resultant Vector):
\[
\text{True Airspeed} = \sqrt{(-378.28)^2 + (28.28)^2} \approx 379.34 \text{ km/hr}
\]

2. **Heading** (Angle of the Resultant Vector):
\[
\theta = \tan^{-1}\left(\frac{28.28}{-378.28}\right)
\]
Since the x-component is negative and the y-component is positive, the angle will be in the second quadrant.
\[
\theta \approx \tan^{-1}\left(-0.0748\right) \approx 175.71^\circ
\]
Thus, the heading is approximately \(175.71^\circ\) from the positive x-axis (East), or \(360^\circ - 175.71^\circ = 184.29^\circ\) from North (which is close to due West with a slight deviation south).

### Final Answer:
- **True Airspeed**: \( \approx 379.34 \) km/hr
- **Heading**: \( \approx 184.29^\circ \) (just south of due West)

Would you like more details on any part of this process? 

Here are five related questions you might find interesting:
1. How would the true airspeed and heading change if the wind speed were higher?
2. What is the significance of true airspeed in aviation navigation?
3. How would the results differ if the wind direction were directly from the north?
4. Can you calculate the wind correction angle required for the airplane to maintain a due west course?
5. How do pilots typically adjust their heading in the presence of crosswinds?

**Tip:** Always break down vectors into components when dealing with problems involving directions and magnitudes!

Learn how to calculate the true airspeed and heading of an airplane affected by wind. This problem involves vector addition and trigonometric calculations to determine the magnitude and direction of the airplane's true airspeed. Suitable for Grades 10-12.

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