An airplane flies due west at an airspeed of 438 mph. The wind blows in the direction of 37° south of west at 52 mph. What is the ground speed of the airplane? What is the bearing of the airplane?
solve bearing in radians

To solve for the ground speed and the bearing of the airplane, we need to break down the problem into components and use vector addition. Here's the step-by-step process:

### 1. Represent the airplane's and wind's velocities as vectors.
- **Airplane velocity vector (\(\vec{V}_a\))**: The airplane is flying due west at 438 mph. 
  \[
  \vec{V}_a = \langle -438, 0 \rangle \, \text{mph}
  \]

- **Wind velocity vector (\(\vec{V}_w\))**: The wind is blowing at 52 mph at an angle of 37° south of west.
  - The x-component (westward): \( V_{wx} = 52 \cos(37^\circ) \)
  - The y-component (southward): \( V_{wy} = -52 \sin(37^\circ) \)
  \[
  \vec{V}_w = \langle 52 \cos(37^\circ), -52 \sin(37^\circ) \rangle \, \text{mph}
  \]

### 2. Calculate the components of the wind vector.
\[
\vec{V}_w = \langle 52 \cos(37^\circ), -52 \sin(37^\circ) \rangle \approx \langle 41.54, -31.34 \rangle \, \text{mph}
\]

### 3. Add the velocity vectors to find the ground speed vector (\(\vec{V}_g\)).
\[
\vec{V}_g = \vec{V}_a + \vec{V}_w = \langle -438, 0 \rangle + \langle 41.54, -31.34 \rangle = \langle -396.46, -31.34 \rangle \, \text{mph}
\]

### 4. Calculate the ground speed (magnitude of \(\vec{V}_g\)).
\[
|\vec{V}_g| = \sqrt{(-396.46)^2 + (-31.34)^2} \, \text{mph}
\]

Let's compute the ground speed.The ground speed of the airplane is approximately **397.7 mph**.

### 5. Calculate the bearing of the airplane.
The bearing is the angle the ground velocity vector makes with the west direction. It is found using the arctangent function:

\[
\theta = \tan^{-1}\left(\frac{V_{gy}}{V_{gx}}\right)
\]

The bearing is measured clockwise from the west direction. We want to convert this bearing into radians.

Let's calculate the bearing in radians.The bearing of the airplane is approximately \(-3.063\) radians. Since this value is negative and measured clockwise from west, it can be interpreted as the angle relative to the negative x-axis (westward direction).

### Summary:
- **Ground Speed**: \( \approx 397.7 \, \text{mph} \)
- **Bearing**: \( \approx -3.063 \, \text{radians} \) (clockwise from west)

Would you like more details or have any questions?

Here are some related questions you might find interesting:
1. How would the ground speed change if the wind speed increased?
2. What would the bearing be in degrees?
3. How does a tailwind affect an airplane's ground speed compared to a headwind?
4. How do you calculate the components of a vector given its magnitude and direction?
5. What effect does wind direction have on an airplane's trajectory?

**Tip:** When working with bearings, it's essential to remember the reference direction and whether the angle is measured clockwise or counterclockwise.

An airplane flies due west at an airspeed of 438 mph. The wind blows in the direction of 37° south of west at 52 mph. What is the ground speed of the airplane? What is the bearing of the airplane? Solve bearing in radians.

Solve the ground speed and bearing of an airplane flying due west at 438 mph, with wind blowing at 52 mph at an angle of 37° south of west. The problem involves vector addition, trigonometry, and converting bearing into radians. Suitable for students in Grades 10-12.

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