Problem 66 - Algebra and Trigonometry with Analytic Geometry

Question

Wind direction and speed An airplane is flying in the direction \(20^{\circ}\) with an airspeed of \(300 \mathrm{mi} / \mathrm{hr}\). Its ground speed and true course are \(350 \mathrm{mi} / \mathrm{hr}\) and \(30^{\circ}\), respectively. Approximate the direction and speed of the wind.

Step-by-Step Solution

Answer

The wind's speed is approximately 68 mi/hr, and its direction is about 87°.

Step 1: Understand the Problem

We need to determine the wind's speed and direction given the airplane's airspeed, ground speed, true course, and the direction the airplane is heading. The airspeed (velocity of the plane relative to the air) is 300 mi/hr at 20°, while the ground speed (velocity relative to the ground) is 350 mi/hr at 30°.

Step 2: Represent the Velocities

Represent the airplane's velocity vectors. - The airspeed vector is directed at 20° with magnitude 300 mi/hr. - The ground speed vector is directed at 30° with magnitude 350 mi/hr. The wind vector can be found by subtracting the airspeed vector from the ground speed vector.

Step 3: Break Down Vectors into Components

Convert the given vectors into their components.- Airplane airspeed vector components: - East (x): \( 300 \cos(20^{\circ}) \) - North (y): \( 300 \sin(20^{\circ}) \)- Airplane ground speed vector components: - East (x): \( 350 \cos(30^{\circ}) \) - North (y): \( 350 \sin(30^{\circ}) \)

Step 4: Calculate Wind Vector Components

Subtract the airspeed vector components from the ground speed vector components to find the wind vector components:- Wind x-component: \( 350 \cos(30^{\circ}) - 300 \cos(20^{\circ}) \)- Wind y-component: \( 350 \sin(30^{\circ}) - 300 \sin(20^{\circ}) \)

Step 5: Determine Wind Speed and Direction

Wind speed is obtained by calculating the magnitude of the wind vector:\[ \text{Wind speed} = \sqrt{(\text{Wind x-component})^2 + (\text{Wind y-component})^2} \]Wind direction is calculated using:\[ \theta = \tan^{-1}\left(\frac{\text{Wind y-component}}{\text{Wind x-component}}\right) \]Convert \(\theta\) from radians to degrees if needed.

Key Concepts

Vectors

Vectors are essential in understanding how objects move in space. They are mathematical objects that have both magnitude and direction. Imagine a vector as an arrow pointing in a certain direction. The length of the arrow represents its magnitude. For example, in the context of flying, vectors can represent the velocity of an airplane or the speed and direction of the wind.
In our exercise, vectors are used to represent both the airplane's airspeed and the ground speed. The airspeed vector points to the direction the plane is heading relative to the air and has a magnitude corresponding to its speed. The ground speed vector represents the plane's motion relative to the ground.
Understanding how these vectors work allows us to calculate important information, such as wind speed and direction.

Angle Measurement

Angles are crucial in trigonometry and vector calculations. They help us understand the direction in which a vector points. Angles are typically measured in degrees or radians.
In our exercise, the airplane has an airspeed angle of 20°, meaning its velocity vector is oriented 20 degrees away from some reference direction. The ground speed angle is given as 30°.
To visualize this, think of a compass where 0° is north, and you rotate clockwise to find the angle. Accurately measuring and interpreting these angles is essential for finding the components of vectors.

Vector Components

Breaking vectors into components makes analyzing them much more manageable. When a vector is described by its components, it's split into two perpendicular parts. Usually, these parts align with the horizontal (x-axis) and vertical (y-axis) directions.
To break a vector using trigonometry:

• The x-component is found using the cosine of the angle: magnitude × cos(angle).
• The y-component is found using the sine of the angle: magnitude × sin(angle).

In our situation, the airplane's airspeed vector is split into east (x) and north (y) components with angles of 20°, and the ground speed vector of 30°.
Once broken into components, you can easily add or subtract them, making it simpler to analyze the situation.

Wind Speed Calculation

To calculate wind speed, you need to consider both the magnitude and direction of the wind vector. This requires using the components calculated from the vectors of the airplane's airspeed and ground speed.
First, subtract the airplane's airspeed vector components from the ground speed components to find the wind's components. This gives you two results: the wind's x-component and y-component.
The magnitude of the wind vector, or wind speed, is found using the Pythagorean theorem:\[\text{Wind speed} = \sqrt{(\text{Wind x-component})^2 + (\text{Wind y-component})^2}\]To find the wind's direction, calculate the angle with the x-axis using the tangent function:\[\theta = \tan^{-1}\left(\frac{\text{Wind y-component}}{\text{Wind x-component}}\right)\]Convert this angle to degrees if it was initially in radians. This direction and speed will give you the complete picture of how the wind is affecting the airplane's course.