Problem 26 - University Physics with Modern Physics

Question

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 \(280 \mathrm{~m}\) at \(30^{\circ}\) east of north. After a fourth displacement,she finds herself back where she started. Use a scale drawing to determine the magnitude and direction of the fourth displacement. (See also Problem 1.61 for a different approach.)

Step-by-Step Solution

Answer

The fourth displacement is approximately 301.1 m at 73.7° south of east.

Step 1: Represent the First Displacement

The first displacement is 180 m straight west. We can represent this vector along the negative x-axis as \[\vec{A} = -180\hat{i} \].

Step 2: Calculate the Components of the Second Displacement

The second displacement is 210 m at an angle of \(45^{\circ}\) east of south. The angle south of the east-west line is \(45^{\circ}\). Thus, the x-component is \[A_{2x} = 210 \cos(135^{\circ}) = 210 \cdot (-\sqrt{2}/2) = -148.5 \]and the y-component is \[A_{2y} = 210 \sin(135^{\circ}) = 210 \cdot (\sqrt{2}/2) = 148.5 \].This vector is represented as \[\vec{B} = (-148.5\hat{i} + 148.5\hat{j})\].

Step 3: Calculate the Components of the Third Displacement

The third displacement is 280 m at an angle of \(30^{\circ}\) east of north. So, the x-component is \[A_{3x} = 280 \cos(30^{\circ}) = 280 \cdot (\sqrt{3}/2) = 242.5 \]and the y-component is \[A_{3y} = 280 \sin(30^{\circ}) = 280 \cdot (1/2) = 140 \].This vector is represented as \[\vec{C} = (242.5\hat{i} + 140\hat{j})\].

Step 4: Calculate the Total Components of the First Three Displacements

Add the x-components and y-components of the three displacements:\[x_{total} = -180 - 148.5 + 242.5 = -86\]and \[y_{total} = 0 + 148.5 + 140 = 288.5\].

Step 5: Determine the Fourth Displacement's Components

Since the spelunker returns to her starting point, the total displacement in both the x and y directions must be zero. Therefore, the x-component and y-component of the fourth displacement are the negatives of the total components of the first three displacements:\[x_{4} = -(-86) = 86\]and \[y_{4} = -(288.5) = -288.5\].Thus, the fourth displacement vector is \[\vec{D} = (86\hat{i} - 288.5\hat{j})\].

Step 6: Compute the Magnitude of the Fourth Displacement

The magnitude of the fourth displacement vector \(\vec{D}\) is given by:\[|\vec{D}| = \sqrt{(86)^2 + (-288.5)^2} = \sqrt{7396 + 83292.25} = \sqrt{90688.25} \approx 301.1\].

Step 7: Determine the Direction of the Fourth Displacement

The direction (angle \(\theta\)) is calculated using the tangent function:\[\theta = \tan^{-1}\left(\frac{-288.5}{86}\right) \approx \tan^{-1}(-3.3547) \approx -73.7^{\circ}\].This angle is measured from the positive x-axis (east direction) toward the negative y-axis (south direction), resulting in a direction of approximately \(73.7^{\circ}\) south of east.

Key Concepts

Magnitude of Vector

When dealing with vector displacement, understanding the magnitude of a vector is crucial. The magnitude, represented often by \(|\vec{D}|\), is simply the length or size of the vector. In the case of the spelunker's fourth displacement, this tells us how far the displacement would take her in a straight line back to her starting point.
To calculate this magnitude, use the Pythagorean theorem, since you are essentially finding the hypotenuse of a right triangle formed by the vector's components. For the fourth displacement vector, which has components \(86\hat{i}\) and \(-288.5\hat{j}\), the magnitude is calculated as:

• Calculate squares: \(86^2 = 7396\) and \((-288.5)^2 = 83292.25\).
• Add these values: \(|\vec{D}| = \sqrt{7396 + 83292.25}\).
• Take the square root to find the magnitude: \(|\vec{D}| \approx 301.1\).

This represents the straightforward distance of her final leg of the journey, back to her original point.

Vector Components

Vector components break a vector into parts that align with the coordinate axes, making calculations easier. Imagine you have a giant arrow pointing somewhere—this arrow can be broken down into smaller, horizontal (x-axis) and vertical (y-axis) arrows. Each of these smaller arrows is a component of the original vector.
For example, in the spelunking exercise, the second displacement vector of 210 meters at an angle 45° east of south is composed of two components:

• The x-component: \(A_{2x} = 210 \cos(135^{\circ}) = -148.5\)
• The y-component: \(A_{2y} = 210 \sin(135^{\circ}) = 148.5\)

This means that move contributes \(-148.5\) meters in the x-direction and \(148.5\) meters in the y-direction.
Breaking vectors up like this allows us to simply add or subtract these smaller vector components to understand the total journey, leading us to determine any missing part, such as the spelunker’s fourth displacement.

Direction of Vectors

Determining a vector's direction involves figuring out the angle it makes with a reference axis, usually the positive x-axis. The direction is crucial for knowing not just how far you are going (the magnitude) but also where you are going, which is especially useful in navigational contexts like spelunking.
The direction of a vector can be calculated using the tangent function, based on the ratio of its y-component to its x-component, expressed as \(\theta = \tan^{-1}(\frac{y}{x})\).
In the exercise, the spelunker’s direction for her fourth displacement, calculated from its components \(86\hat{i}\) and \(-288.5\hat{j}\), is:

• Use the formula: \( heta = \tan^{-1}\left(\frac{-288.5}{86}\right)\).
• The result: \(-73.7^{\circ}\), which means 73.7° south of east.

Here, the negative sign suggests a direction away from the positive side of the x-axis, which helps in mapping her exact final path back to the start.