Problem 43 - College Physics

Question

A plane leaves Seattle, flies \(85 \mathrm{mi}\) at \(22^{\circ}\) north of east, and then changes direction to \(48^{\circ}\) south of east. After flying \(115 \mathrm{mi}\) in this new direction, the pilot must make an emergency landing on a field. The Seattle airport facility dispatches a rescue crew. (a) In what direction and how far should the crew fly to go directly to the field? Use components to solve this problem. (b) Check the reasonableness of your answer with a careful graphical sum.

Step-by-Step Solution

Answer

The rescue crew should fly 164.8 miles at 19.1° south of east.

Step 1: Breakdown the first leg of the flight

The plane first flies 85 miles at an angle of \( 22^\circ \) north of east. We need to find the east (X) and north (Y) components.- East component (X1) = \( 85 \cos(22^\circ) \)- North component (Y1) = \( 85 \sin(22^\circ) \)

Step 2: Calculate first leg components

Calculate:- \( X1 = 85 \cos(22^\circ) \approx 78.86 \) miles- \( Y1 = 85 \sin(22^\circ) \approx 31.80 \) miles

Step 3: Breakdown the second leg of the flight

The plane then flies 115 miles at an angle of \( 48^\circ \) south of east. We need to find the east (X) and south (Y) components.- East component (X2) = \( 115 \cos(48^\circ) \)- South component (Y2) = \( 115 \sin(48^\circ) \)

Step 4: Calculate second leg components

Calculate:- \( X2 = 115 \cos(48^\circ) \approx 77.11 \) miles- \( Y2 = 115 \sin(48^\circ) \approx 85.76 \) miles (note, this is south, so it will be a negative value when summing).

Step 5: Sum the components

Add the components from both legs to find the plane's final position relative to Seattle:- Total East = \( X1 + X2 = 78.86 + 77.11 \approx 155.97 \) miles- Total North = \( Y1 - Y2 = 31.80 - 85.76 \approx -53.96 \) miles (negative indicates south).

Step 6: Calculate the rescue crew's direction and distance

The rescue crew should fly directly to the field along the resultant of these vectors.- Distance (R) is calculated using the Pythagorean theorem: \[ R = \sqrt{(155.97)^2 + (-53.96)^2} \approx 164.8 \text{ miles} \]- The angle \( \theta \) south of east is given by: \[ \theta = \tan^{-1}\left(\frac{|53.96|}{155.97}\right) \approx 19.1^\circ \text{ south of east} \]

Step 7: Verify with a graphical sum

Draw a vector diagram to verify:- The first vector 85 miles at \( 22^\circ \) north of east.- The second vector 115 miles at \( 48^\circ \) south of east.- Closing the triangle with the resultant, which should show alignment with calculated direction and distance.

Key Concepts

Trigonometry

Trigonometry is a branch of mathematics that helps us understand the relationships between angles and distances in triangles. In this problem, trigonometry is crucial to determine how far and in what direction a rescue crew should fly to reach a plane's emergency landing site. Specifically, we use trigonometric functions such as sine and cosine to break down vector paths into components.

When a plane flies at a particular angle and distance, it forms a triangle with the horizontal (east-west) and vertical (north-south) axes.

• The cosine function helps calculate the horizontal distance (east-west axis) covered: - If a plane travels 85 miles at a 22° angle north of east, the east component is found using cosine: \(85 \cos(22°)\).
• Similarly, the sine function calculates the vertical distance (north-south axis) covered: - Here, it's north since the direction is north of east: \(85 \sin(22°)\).

Trigonometry plays a similar role when the direction changes, such as flying 115 miles at 48° south of east. By using these functions, we decompose the entire journey into manageable directional components.

Graphical Representation of Vectors

Vectors are helpful tools in physics and navigation for representing direction and magnitude. A vector can showcase both where something is going and how far it travels. In our scenario, vectors symbolize the path and direction of the plane.

Graphically representing vectors involves drawing them as arrows on a coordinate system, where their length is proportionate to their magnitude, and their direction matches the traveled path's direction. You can draw:

• An arrow representing the first leg of the journey (85 miles, 22° north of east). Its eastward part can be seen with a horizontal arrow, while its upward component (north) is shown vertically.
• A second arrow for the next segment (115 miles, 48° south of east), pointing mostly east but downward since it's south.

To find where the plane landed, you add these vectors. This involves aligning them tail to head—starting the second arrow where the first one ends. The missing side—connecting the start of the first vector and the end of the second—shows the direct path the rescue crew must follow. A graphical sum provides a tangible way to verify your mathematical calculations.

Emergency Rescue Navigation

In emergencies, accurately navigating to a plane's landing site is essential for timely rescue. Emergency rescue navigation involves calculating the shortest, most efficient route from point A to B. This ensures that rescue efforts are swift and accurate.

In this exercise, once the plane's final position is calculated using vectors, trigonometry helps identify the direct path. To guide a rescue crew effectively:

• Combine the east and southbound components of the plane’s path into a single resultant vector.
• Use the Pythagorean theorem to find the distance the rescue crew must fly: - Here, it's 164.8 miles.
• Determine the direction with the tangent function to calculate the angle relative to south of east. - In this case, around 19.1° south of east.

This navigation method guarantees the crew flies directly to where the plane landed, optimizing both time and resources. It also underscores the importance of mathematical precision and clarity in high-stakes scenarios like emergency rescues.