Problem 61 - University Physics with Modern Physics

Question

As noted in Exercise 1.26, a spelunker is surveying a cave. She follows a passage 180 m straight west, then 210 m in a direction 45\(^{\circ}\) east of south, and then 280 m at 30\(^{\circ}\) east of north. After a fourth displacement, she finds herself back where she started. Use the method of components to determine the magnitude and direction of the fourth displacement. Draw the vector-addition diagram and show that it is in qualitative agreement with your numerical solution.

Step-by-Step Solution

Answer

The fourth displacement is approximately 211.2 m, directed 2.31° south of west.

Step 1: Understanding the Path

The spelunker moves in three distinct paths:1. 180 m west2. 210 m at 45\(^{\circ}\) east of south3. 280 m at 30\(^{\circ}\) east of north Now we need to calculate the fourth displacement that will bring the spelunker back to the starting point using vector addition.

Step 2: Representing Each Path as Components

Represent each part of the path in terms of its vector components:1. **West Move:** - **x-component:** \(-180\, \text{m} \) - **y-component:** \(0\, \text{m} \)2. **45\(^{\circ}\) East of South:** (210 m) - **x-component:** \(210 \cdot \cos(45^{\circ}) \) - **y-component:** \(-210 \cdot \sin(45^{\circ}) \)3. **30\(^{\circ}\) East of North:** (280 m) - **x-component:** \(280 \cdot \cos(30^{\circ}) \) - **y-component:** \(280 \cdot \sin(30^{\circ}) \)

Step 3: Calculating the Components

Calculate the x and y components for each part of the path:1. **First Displacement (West):** - \(x_1 = -180\, \text{m} \) - \(y_1 = 0\, \text{m} \)2. **Second Displacement (45\(^{\circ}\) East of South):** - \(x_2 = 210 \cdot \frac{\sqrt{2}}{2} = 148.5\, \text{m} \) - \(y_2 = -210 \cdot \frac{\sqrt{2}}{2} = -148.5\, \text{m} \)3. **Third Displacement (30\(^{\circ}\) East of North):** - \(x_3 = 280 \cdot \frac{\sqrt{3}}{2} = 242.5\, \text{m} \) - \(y_3 = 280 \cdot \frac{1}{2} = 140\, \text{m} \)

Step 4: Sum of Components

Sum up the x-components and y-components to determine the resultant components:- **Total x-component:** \[ x_{\text{total}} = x_1 + x_2 + x_3 = -180 + 148.5 + 242.5 = 211\, \text{m} \]- **Total y-component:** \[ y_{\text{total}} = y_1 + y_2 + y_3 = 0 - 148.5 + 140 = -8.5\, \text{m} \]

Step 5: Calculate Fourth Displacement Components

Since the spelunker returns to the starting point, the sum of all displacements must equal zero:- The x-component of the fourth displacement, \(x_4\), must satisfy \(x_{\text{total}} + x_4 = 0\): \(-211 + x_4 = 0\) implies \(x_4 = -211\, \text{m} \)- The y-component of the fourth displacement, \(y_4\), must satisfy \(y_{\text{total}} + y_4 = 0\): \(-8.5 + y_4 = 0\) implies \(y_4 = 8.5\, \text{m} \)

Step 6: Calculate Magnitude and Direction of the Fourth Displacement

Determine the magnitude \(d_4\) and direction \(\theta\) of the fourth displacement:- **Magnitude:** \[ d_4 = \sqrt{x_4^2 + y_4^2} = \sqrt{(-211)^2 + (8.5)^2} \approx 211.2\, \text{m} \]- **Direction:** \( \theta = \tan^{-1}\left(\frac{y_4}{x_4}\right) = \tan^{-1}\left(\frac{8.5}{-211}\right) \approx -2.31^{\circ}\) relative to the negative x-axis. This results in roughly \(2.31^{\circ}\) south of west.

Key Concepts

Displacement

Displacement is a vector quantity that represents the change in position of an object. It is defined as the shortest straight line between the initial and final positions, making it different from distance, which is the total path traveled. In this exercise, displacement plays a crucial role as we are tasked with finding the fourth displacement necessary for the spelunker to return to her starting point.

It's important to remember that displacement is direction-aware, meaning it considers where the movement is happening, not just how far the object has moved. For instance, going 180 m west is different from going 180 m east because of the opposite directions. To solve problems involving displacement, we often break down the vector into its components along the x (horizontal) and y (vertical) axes, which makes computations much simpler.

Components of Vectors

When tackling complex vector problems like finding the fourth displacement here, it's useful to break vectors into their components. Each vector has two components: the horizontal (x-component) and the vertical (y-component). By doing this, we can use simple trigonometric identities to handle each part separately.

• The x-component (horizontal) is usually calculated using the cosine of the given angle.
• The y-component (vertical) is found using the sine of the given angle.

This method makes calculations more manageable, since you can add the x-components and y-components independently to find the resultant vector. In the provided exercise, we neatly calculated the individual x and y components for each displacement. Consequently, we could directly apply these sums to figure out what the final x and y components need to be for her to return to the starting point.

Trigonometry in Physics

Trigonometry is invaluable in physics, especially when dealing with vector operations, like in this cave exploration scenario. We use trigonometric functions, such as sine and cosine, to resolve vectors into their components. Moreover, the tangent function is employed to determine the angle of a vector from its components.

For calculating vector components:

• Cosine relates to the x-component: \(x = ext{magnitude} \cdot \cos( heta)\)
• Sine refers to the y-component: \(y = ext{magnitude} \cdot \sin( heta)\)

Finally, to find the angle for our resultant vector, we used the inverse tangent function, which finds the angle with respect to the x-axis: \(\theta = \tan^{-1}\left(\frac{y}{x}\right)\). This calculation helps determine the vector's direction, crucial for understanding the physical movement relative to compass directions or axes. In our example, we quantified the angle relative to the negative x-axis, giving us the exact direction needed for the spelunker to get back to her start point.