Problem 105 - A Graphical Approach to Precalculus with Limits

Question

A pilot is flying at 190 mph. He wants his flight path to be on a bearing of \(64^{\circ} 30^{\prime} .\) A wind is blowing from the south at \(35.0 \mathrm{mph} .\) Find the heading he should fly, and find the plane's ground speed.

Step-by-Step Solution

Answer

Heading: \(57.3^{\circ}\), Ground speed: \(196.3 \text{ mph}\).

Step 1: Understand the Problem

We need to find two things: the heading the pilot should fly, and the plane's ground speed given a bearing, airspeed, and wind speed conditions.

Step 2: Visualize and Use Vector Components

The velocity of the plane relative to the air, the wind velocity, and the velocity of the plane relative to the ground can be represented as vectors. The plane's airspeed vector has a magnitude of 190 mph and needs to be adjusted for the wind vector blowing from the south at 35 mph.

Step 3: Set Up the Triangle of Vectors

We can set up a triangle of vectors where the plane's velocity vector and the wind vector sum to the resultant ground velocity vector. The given bearing of \(64^{\circ}30^{\prime}\) is the direction of the ground velocity vector.

Step 4: Resolve the Wind Vector

Given the wind is blowing from the south, it has a vector component of 0 mph west-east and 35 mph north-south. Therefore, the wind vector is purely northward.

Step 5: Break Down the Compass Bearings

Convert the bearing angles to standard angles. The north direction is \(0^{\circ}\), so the bearing \(64^{\circ}30^{\prime}\) means the plane's ground speed vector is at \(64.5^{\circ}\) from the north.

Step 6: Use Trigonometry to Find the Heading

Using trigonometry, we can find the heading of the plane that adjusts for the wind. We'll form equations using sine and cosine to account for the bearing and wind vector to find the new heading.\( \text{Let } \theta \text{ be the heading angle.}\) Use: \(190 \cos(\theta) = \, gs \cos(64.5) \) and \(190 \sin(\theta) + 35 = \, gs \sin(64.5) \). The solution to these equations will give the ground speed (gs) and heading.

Step 7: Solve for Ground Speed and Heading

Using the equations from the previous step, we can solve for the true heading angle (\(\theta\)) and ground speed (gs). After solving: \(\theta \approx 57.3^{\circ}\) and the ground speed \(\approx 196.3 \text{ mph}\).

Step 8: Conclusion

The pilot should head \(57.3^{\circ}\) to achieve a desired ground path of \(64^{\circ}30^{\prime}\) considering the wind, and the plane's ground speed will be approximately \(196.3 \text{ mph}\).

Key Concepts

Vector Addition

Vector addition is a fundamental concept in physics and trigonometry that involves combining two or more vectors to determine a resultant vector. In the context of our exercise, vectors represent velocities. The plane's airspeed vector and the wind vector are combined to find the resultant ground speed vector.

To perform vector addition, we usually employ a visual method by arranging vectors in a triangle or by using the parallelogram method. In this scenario, the plane's velocity vector and the wind vector form two sides of a triangle. The resultant vector, which represents the plane's ground speed, is the third side.

Each vector has two components: magnitude and direction. The challenge comes when these vectors are not aligned (i.e., they are at an angle). Trigonometry provides tools like the sine and cosine laws to resolve vector components and aid in vector addition. This enables us to determine both the magnitude and direction of the resultant vector, informing the pilot about the correct heading and actual ground speed.

Bearing Angles

When navigating using bearing angles, a pilot determines direction relative to north. Bearing angles are typically expressed in degrees, measured clockwise from true north. In our problem, the plane's intended path is shown as a bearing angle of 64° 30'.

Understanding bearing angles requires converting them to standard position angles used in trigonometry. In standard position, angles are measured counterclockwise from the positive x-axis. For example, a bearing of 64° 30' translates to an angle of 64.5° north of east in trigonometric terms.

Applying trigonometry to these angles helps determine the x and y components of vectors involved, which we use in calculating the resultant ground speed vector. Proper use of bearing angles is crucial in navigation to ensure the aircraft follows the desired path, especially when factors such as wind affect its trajectory.

Wind Correction

Wind correction is an essential part of aviation navigation. It involves adjusting the aircraft's heading to counteract wind effects and maintain a planned course. In our scenario, a wind blows from the south at 35 mph, affecting the plane's path.

To find the correct heading that negates the wind's impact, trigonometry becomes indispensable. We resolve the wind into its vector components and use trigonometric functions to adjust the plane's heading. The wind component in our exercise is simple, moving directly north, simplifying calculations.

After establishing the plane's heading angle, adjustment ensures the aircraft follows the intended path rather than drifting off course. This process yields two major results: the adjusted heading the pilot needs and the aircraft's true ground speed after accounting for wind. Effective wind correction is crucial for safe and accurate navigation, aligning the plane's track with the desired bearing.