Problem 70 - University Physics with Modern Physics

Question

Emergency Landing. A plane leaves the airport in Galisteo and flies 170 \(\mathrm{km}\) at \(68^{\circ}\) east of north and then changes direction to fly 230 \(\mathrm{km}\) at \(48^{\circ}\) south of east, after which it makes an immediate emergency landing in a pasture. When the airport sends out a rescue crew, in which direction and how far should this crew fly to go directly to this plane?

Step-by-Step Solution

Answer

The rescue crew should fly 329.3 km in the direction 19° south of east.

Step 1: Break Down the Problem

We want to find a direct path (magnitude and direction) to the plane from the starting point. The plane's path involves two segments: 170 km at 68° east of north, and 230 km at 48° south of east. We'll use vector addition to combine these two paths and find the resultant vector from the starting point to the plane's final position.

Step 2: Define the Coordinate System

Let positive y represent north and positive x represent east. We will decompose each segment of the journey into x and y components.

Step 3: Decompose the Vector Segments

For the first segment: 170 km at 68° east of north, \(x_1 = 170 \cdot \sin(68°)\),\(y_1 = 170 \cdot \cos(68°)\).For the second segment: 230 km at 48° south of east, \(x_2 = 230 \cdot \cos(48°)\),\(y_2 = -230 \cdot \sin(48°)\) because it's south of east.

Step 4: Calculate the Components

Start by computing each component:\(x_1 = 170 \cdot 0.9272 \approx 157.6\) km,\(y_1 = 170 \cdot 0.3746 \approx 63.7\) km,\(x_2 = 230 \cdot 0.6691 \approx 153.9\) km,\(y_2 = -230 \cdot 0.7431 \approx -171.0\) km.

Step 5: Sum the Components

Add the x-components and y-components respectively:\(x_{total} = x_1 + x_2 = 157.6 + 153.9 = 311.5\) km, \(y_{total} = y_1 + y_2 = 63.7 - 171.0 = -107.3\) km.

Step 6: Calculate the Resultant Magnitude

Use the Pythagorean theorem to find the magnitude of the resultant vector:\[ R = \sqrt{x_{total}^2 + y_{total}^2} = \sqrt{(311.5)^2 + (-107.3)^2} \approx 329.3 \text{ km} \].

Step 7: Find the Direction of the Resultant

Calculate the angle \( \theta \) from the east direction using \( \tan \theta = \frac{|y|}{x} = \frac{107.3}{311.5} \). Thus, \( \theta = \tan^{-1}(0.344) \approx 19.0^{\circ} \). Since the resultant vector is in the southeast quadrant, the direction is 19° south of east.

Key Concepts

Coordinate System

When working with vector quantities in physics, it's important to use a coordinate system to determine direction and magnitude precisely. A coordinate system typically involves an origin, usually marked as the point (0,0), from which you can define positions using axes. For problems involving navigation, a common choice is:

• The positive x-axis: representing east.
• The positive y-axis: representing north.

You can think about this as placing a map flat and determining directions based on compass points. This setup helps to accurately resolve vector components into manageable parts, further aiding calculations, especially when dealing with trigonometric functions.

Vector Components

Understanding vector components is crucial when applying vectors in physics, as each vector can be split into two parts that align with the coordinate axes. By breaking down a vector into its components, we can handle them like simple numbers.

• The x-component represents the projection onto the east-west axis.
• The y-component represents the projection onto the north-south axis.

For example, in vector addition problems, a vector traveling a certain distance at an angle needs to be decomposed into its respective components. Using trigonometric functions such as sine and cosine, we can find these components, offering a clearer way to continue with calculations.

Pythagorean Theorem

The Pythagorean theorem is a foundational principle in mathematics that is often used in physics when dealing with vectors. It relates the sides of a right triangle, making it perfect for determining the magnitude of a resultant vector in a two-dimensional plane.When you have the components of a vector, you can use the formula:\[ R = \sqrt{x^2 + y^2} \]Where \( R \) is the hypotenuse or the resultant vector, \( x \) is the length along the x-axis, and \( y \) is the length along the y-axis. This gives the straight-line distance from the starting point to the endpoint, crucial in navigation problems, such as finding a direct route to a location.

Trigonometry in Physics

Trigonometry is a powerful tool in physics, especially when resolving vectors and calculating angles. It involves functions like sine, cosine, and tangent, which relate the angles and sides of a triangle.

• Cosine is used to find the adjacent side of an angle in a right triangle, which corresponds to the x-component.
• Sine is used to find the opposite side, which corresponds to the y-component.

To determine the direction of a vector, the tangent function is often used. By calculating:\[ \tan \theta = \frac{|y|}{x} \]We find the angle \( \theta \) related to the adjacent side, giving insight into the vector's orientation regarding the coordinate system. When dealing with navigation or physics problems, understanding this relationship is key to finding both direction and accurately representing movement.