Problem 70 - Contemporary Precalculus

Question

A plane is flying in the direction \(200^{\circ}\) with an air speed of \(500 \mathrm{mph}\). Its course and ground speed are \(210^{\circ}\) and \(450 \mathrm{mph}\) respectively. What are the direction and speed of the wind?

Step-by-Step Solution

Answer

Answer: The approximate wind speed is 52.46 mph, and the direction is 239.56 degrees.

Step 1: Understand the relationship between air speed, ground speed, and wind speed

To find the direction and speed of the wind, we need to understand the relationship between the plane's air speed, ground speed, and wind speed. The ground speed is the resulting speed of the plane, which is the vector addition of its air speed and wind speed.

Step 2: Represent the given information as vectors

We represent the given information as vectors. Let \(V_p\) be the plane's air speed vector, \(V_g\) be the ground speed vector and \(V_w\) be the wind speed vector. The given information is: - \(V_p = 500 \mathrm{mph}\) at \(200^{\circ}\) - \(V_g = 450 \mathrm{mph}\) at \(210^{\circ}\) Since the ground speed \(V_g\) is the vector addition of air speed \(V_p\) and wind speed \(V_w\), we have: \(V_g = V_p + V_w\) Now, we need to find \(V_w\).

Step 3: Write given vectors in component form

To solve for \(V_w\), we need to write the given vectors in component form. We can find the x and y components of \(V_p\) and \(V_g\) using trigonometric functions: \(V_p = (500\cos(200^{\circ}), 500\sin(200^{\circ}))\) \(V_g = (450\cos(210^{\circ}), 450\sin(210^{\circ}))\)

Step 4: Calculate the wind speed vector

Since \(V_g = V_p + V_w\), we can rewrite the equation to find \(V_w\): \(V_w = V_g - V_p\) Substituting the component form of \(V_p\) and \(V_g\), we get: \(V_w = (450\cos(210^{\circ}) - 500\cos(200^{\circ}), 450\sin(210^{\circ}) - 500\sin(200^{\circ}))\)

Step 5: Calculate the magnitude and direction of the wind vector

To find the speed of the wind, we need to find the magnitude of the wind vector \(V_w\). We can use the Pythagorean theorem to find the magnitude of \(V_w\): \(|V_w| = \sqrt{[(450\cos(210^{\circ}) - 500\cos(200^{\circ}))^2 + (450\sin(210^{\circ}) - 500\sin(200^{\circ}))^2]}\) To find the direction of the wind, we can use the arctangent function: \(\theta = \arctan \dfrac{450\sin(210^{\circ}) - 500\sin(200^{\circ})}{450\cos(210^{\circ}) - 500\cos(200^{\circ})}\)

Step 6: Simplify and find the final answer

Using a calculator, we can now compute the exact value for the magnitude and direction of the wind vector: \(|V_w| \approx 52.46 \mathrm{mph}\) \(\theta \approx 239.56^{\circ}\) Therefore, the wind speed is approximately \(52.46 \mathrm{mph}\) and the direction is approximately \(239.56^{\circ}\).

Key Concepts

Air Speed and Ground Speed

Air speed and ground speed are fundamental concepts in aviation. Air speed is the speed of an aircraft relative to the air through which it moves. Ground speed, on the other hand, is the speed of the aircraft relative to the ground.
Different from air speed, ground speed accounts for influences like wind. These speeds can be represented as vectors, where both magnitude and direction are considered.
This means air speed is one vector pointing in a particular direction, and the wind can alter this by adding another vector. Therefore, the plane's ground speed is often the result of the vector addition of the air speed and wind speed vectors.

Wind Speed Calculation

To calculate wind speed and direction, we consider how wind affects an aircraft's course and speed. We understand these factors using vector mathematics. The exercise demonstrates this by initially representing air speed and ground speed as vectors. Each vector has a magnitude, representing speed in miles per hour, and a direction in degrees.
Wind speed calculation involves finding the difference between ground speed and air speed vectors.

• We use trigonometric functions to decompose each vector into its x and y components.
• This decomposition helps represent the precise effect of wind.
• By rearranging the vector equation, we solve for the wind's x and y components.

Once identified, these components allow us to calculate both the magnitude (using the Pythagorean theorem) and direction (using the arctangent function) of the wind vector. The magnitude gives us the wind speed, and the direction tells us where the wind is coming from in degrees.

Magnitude and Direction of Vectors

Understanding vectors involves knowing their magnitude and direction. Magnitude signifies the length or size of the vector, akin to speed in physical terms. Direction specifies the orientation of the vector in a plane, usually given in degrees.
Calculating the magnitude of a vector involves squaring its components, summing them, and then taking the square root \( |\mathbf{v}| = \sqrt{x^2 + y^2} \).
Assessing the direction uses the arctangent of the ratio of the y-component to the x-component \( \theta = \arctan\left(\frac{y}{x}\right) \). These calculations allow us to fully understand a vector's impact and contribution in scenarios, such as determining wind speed's influence on a plane's flight path.

• Knowing how to change between polar coordinates (magnitude, direction) and rectangular coordinates (x, y components) is essential.
• This skill is critical in various fields, such as navigation and engineering, where vector manipulation is vital to accurate problem-solving.

Mastery of magnitude and direction allows one to efficiently manage and predict vector influences in any vector-related task.