Problem 101 - A Graphical Approach to Precalculus with Limits

Question

An airline route from San Francisco to Honolulu is on a bearing of \(233.0^{\circ} .\) A jet flying at 450 mph with that heading runs into a wind blowing at 39.0 mph from a direction of \(114.0^{\circ} .\) Find the final bearing and ground speed of the plane.

Step-by-Step Solution

Answer

Ground speed: ~445 mph; Bearing: ~230°.

Step 1: Understand the Scenario

The plane is flying from San Francisco to Honolulu at a bearing of \(233.0^{\circ}\). This means it's moving southwest. The wind is blowing at \(39\) mph from \(114.0^{\circ}\), which is roughly east-southeast.

Step 2: Convert Wind Direction to Bearing Format

Since the wind is blowing from \(114.0^{\circ}\), it moves towards \(294.0^{\circ}\). This involves adding \(180\) degrees to the direction if it's less than \(180.0\) degrees.

Step 3: Break Down Velocities into Components

The plane's velocity components are calculated using its speed and direction: \[ v_{plane,x} = 450 \cos(233.0^{\circ}) \] \[ v_{plane,y} = 450 \sin(233.0^{\circ}) \] For the wind: \[ v_{wind,x} = 39 \cos(294.0^{\circ}) \] \[ v_{wind,y} = 39 \sin(294.0^{\circ}) \]

Step 4: Calculate Resultant Velocity Components

Add the plane's and wind's components to find the resultant velocity components: \[ v_{resultant,x} = v_{plane,x} + v_{wind,x} \] \[ v_{resultant,y} = v_{plane,y} + v_{wind,y} \]

Step 5: Determine Ground Speed

Use the Pythagorean theorem to find the ground speed (magnitude of resultant velocity): \[ \text{Ground Speed} = \sqrt{v_{resultant,x}^2 + v_{resultant,y}^2} \]

Step 6: Calculate Final Bearing

Calculate the angle of the resultant vector using the arctangent function: \[ \theta = \tan^{-1}\left(\frac{v_{resultant,y}}{v_{resultant,x}}\right) \] Adjust \(\theta\) for proper quadrant location if needed and convert it into a bearing.

Key Concepts

Understanding Bearing in Navigation

Bearing is a direction or path along which something moves or along which it lies. In navigation, it's crucial because it helps pilots, sailors, and others orient themselves. A bearing is usually measured in degrees. For example, true north is expressed as 0 degrees.
When a plane runs on a certain bearing, it follows a straight path aligned with the compass directions in degrees. In the given exercise, the plane heads on a bearing of $233.0^{ ext{°}}$, which indicates a southwest direction. Southwest falls between south (180°) and west (270°) on the compass.
What's important in navigation is the distinction between 'from' and 'toward' directions. If the wind comes from 114°, it moves towards 294°. Understanding these directions helps convert wind issues into a manageable plane path.

Calculating Ground Speed

Ground speed is the speed of an aircraft relative to the ground. It is not always identical to the airspeed (speed through the air) due to wind conditions.
To find the ground speed when the wind affects a plane's flight, we break down the velocities into components along the x-axis (east-west) and y-axis (north-south). Calculating these components accurately will aid in deriving the ground speed using the Pythagorean theorem:

• For the plane’s path: \[v_{plane,x} = 450 \cos(233.0^{\circ})\]\[v_{plane,y} = 450 \sin(233.0^{\circ})\]
• Describing the wind: \[v_{wind,x} = 39 \cos(294.0^{\circ})\]\[v_{wind,y} = 39 \sin(294.0^{\circ})\]

Combining these results aids in calculating the resultant velocity components, and thus, the ground speed:
\[\text{Ground Speed} = \sqrt{v_{resultant,x}^2 + v_{resultant,y}^2}\]This formula retrieves the actual speed at which the plane travels over the ground.

Determining Velocity Components

Vector addition helps discover the effect of two combined velocity influences: the plane's own speed and the speed of the wind. We're interested in finding out how these influences change the final course and speed of the craft.
When breaking the velocity into components, you'd start by determining how much of the velocity acts horizontally (x-component) and vertically (y-component). This is accomplished by using trigonometric functions like cosine and sine, applying them with the direction angle.
Such vector components allow accurate real-world scenarios for navigation tasks and help in computing the resultant or effective velocity. Without knowing each part of the vector's pull in its specific direction, it would be challenging to accurately predict or plan paths in navigation.