Problem 6 - College Physics

Question

A student walks \(100 \mathrm{~m}\) west and \(50 \mathrm{~m}\) south. (a) To get back to the starting point, the student must walk in a general direction of (1) south of west, (2) north of east, (3) south of east, (4) north of west. (b) What displacement will bring the student back to the starting point?

Step-by-Step Solution

Answer

(a) Direction: north of east. (b) Displacement: approximately 111.8 meters.

Step 1: Understand the Problem

The student first walks 100 meters west, followed by 50 meters south. We need to determine the direction and displacement required for the student to return to the starting point.

Step 2: Determine the Return Direction

To return to the starting point from where the student is after walking west and south, they need to head in the opposite direction of their movements. Walking west and south results in needing to walk east and north to return. Thus, the correct direction will be north of east.

Step 3: Calculate the Displacement Using Pythagorean Theorem

The displacement is the straight line distance from the final position back to the starting point. Since the movements of 100 meters west and 50 meters south form a right triangle, we apply the Pythagorean theorem:\[ c = \sqrt{(100)^2 + (50)^2} \]Calculating this gives:\[ c = \sqrt{10000 + 2500} = \sqrt{12500} = 50\sqrt{5} \approx 111.8 \text{ meters} \]

Step 4: Conclude the Displacement

The calculated displacement value means the student needs to walk approximately 111.8 meters in a direction north of east to return to the starting point.

Key Concepts

Pythagorean theorem

When dealing with problems involving distances in two dimensions, the Pythagorean theorem is an essential tool. It allows you to find the length of the "hypotenuse" in a right-angled triangle, which is the straight-line distance or displacement in many practical problems. The theorem states:

\[ c = \sqrt{a^2 + b^2} \]

Here, \(c\) is the hypotenuse, and \(a\) and \(b\) are the lengths of the other two sides of the triangle. For our problem, when the student walks 100 meters west (one side of the triangle) and 50 meters south (the other side), these distances form a right angle with each other. By calculating \(c\), we find the shortest path—or displacement—the student needs to return. Breaking down in simple steps:

• Square each distance: \(100^2\) (west) and \(50^2\) (south).
• Add the squared values: \(10000 + 2500 = 12500\).
• Take the square root: \(\sqrt{12500} \approx 111.8\).

This gives us the straight-line distance the student must travel to head back to the starting point.

Directional movement

Understanding directional movement is crucial when determining how to get from one point to another, especially when dealing with compass directions. Initially, the student moves west and south. To return to the starting point, the opposite movements are required: north and east. Hence, the logical return path involves moving north of east.

Unlike cardinal directions (N, S, E, W), the phrase "north of east" suggests a mix of two directions. This means the student doesn’t move pure north or east but rather at an angle that combines both. Here's how you can think of it:

• Visualize moving straight east, then start moving north from this line.
• You'd be moving more north than east relative to the original eastward path.

This {north} of {east} methodology provides a clear, combined compass direction for practical navigation.

2D motion

2D motion introduces the concept of movements occurring on a flat plane, rather than a single line. It's like moving on a map, where you can go north-south (latitude) and east-west (longitude). When calculating motions and displacements, each movement vector is crucial. In this problem:

• The student walks west: a horizontal movement on the x-axis.
• Then walks south: a vertical movement on the y-axis.

The combination of these creates a right triangle on a 2D plane. When the student wants to return to the starting point, they need to think in vectors: the total horizontal and vertical movements must cancel out. By using the Pythagorean theorem, we're calculating the displacement vector, which represents the direct path back. This principle is foundational in physics and navigation, as any calculated movement can be plotted as coordinates or vectors on a 2D system.