The captain of a plane wishes to proceed due west. The cruising speed of the plane is 245 m/s relative to the air. A weather report indicates that a 38.0 m/s wind is blowing from the south to the north. In what direction, measured with respect to due west, should the pilot head the plane relative to the air?

Options

A. `81.1^(@)` south of west
B. `17.8^(@)` south of west
C. `8.92^(@)` south of west
D. `8.82^(@)` north of west

Answer

To solve the problem, we need to determine the direction in which the pilot should head the plane relative to the air to ensure that the plane travels due west, taking into account the wind blowing from south to north.

### Step-by-Step Solution:

1. **Identify the Given Information:**
- Cruising speed of the plane (V_p) = 245 m/s (relative to air)
- Wind speed (V_w) = 38 m/s (from south to north)

2. **Set Up the Coordinate System:**
- Let the west direction be negative x-axis (-i).
- Let the north direction be positive y-axis (+j).
- Therefore, the wind vector can be represented as:
\[ \vec{V_w} = 0 \hat{i} + 38 \hat{j} \text{ m/s} \]

3. **Determine the Required Ground Velocity:**
- The desired velocity of the plane relative to the ground (V_g) is:
\[ \vec{V_g} = -V_p \hat{i} + 0 \hat{j} \text{ m/s} \]
- This means the plane needs to travel due west with no north-south component.

4. **Express the Velocity of the Plane Relative to Air:**
- Let the angle the pilot should head the plane be θ with respect to due west. The velocity of the plane relative to the air can be expressed as:
\[ \vec{V_p} = V_p \cos(\theta) \hat{i} + V_p \sin(\theta) \hat{j} \]
- Here, \( V_p \cos(\theta) \) is the westward component and \( V_p \sin(\theta) \) is the northward component.

5. **Set Up the Equation for Ground Velocity:**
- The ground velocity can be expressed as:
\[ \vec{V_g} = \vec{V_p} + \vec{V_w} \]
- Therefore:
\[ -V_p \hat{i} + 0 \hat{j} = (V_p \cos(\theta) \hat{i} + V_p \sin(\theta) \hat{j}) + (0 \hat{i} + 38 \hat{j}) \]

6. **Separate the Components:**
- For the x-component:
\[ -V_p = V_p \cos(\theta) \]
- For the y-component:
\[ 0 = V_p \sin(\theta) + 38 \]

7. **Solve for the Components:**
- From the y-component equation:
\[ V_p \sin(\theta) = -38 \]
\[ \sin(\theta) = -\frac{38}{245} \]

- From the x-component equation:
\[ -V_p = V_p \cos(\theta) \implies \cos(\theta) = -1 \]

8. **Calculate the Angle θ:**
- Now, we can find θ using the sine function:
\[ \theta = \sin^{-1}\left(-\frac{38}{245}\right) \]
- Calculating this gives:
\[ \theta \approx -8.92^\circ \]
- This means the pilot should head approximately 8.92 degrees south of west.

### Final Answer:
The pilot should head the plane approximately **8.92 degrees south of west**.