Distance and Displacement - Physics

1. Fundamental Concepts

Definition: Distance is the total length of the path traveled between two points, regardless of direction.
Displacement: Displacement is the shortest distance from the initial to the final position with a specified direction.
Scalar vs. Vector: Distance is a scalar quantity (magnitude only), while displacement is a vector quantity (magnitude and direction).

2. Key Concepts

Basic Rule: $${\text{{Distance}}} \geq {\text{{Displacement}}}$$

Calculation: $${\text{{Displacement}}} = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$$

Application: Used in navigation, physics experiments, and everyday life for understanding motion

3. Examples

Example 1 (Basic)

Problem: A person walks 3 meters east and then 4 meters north. What is the displacement?

Step-by-Step Solution:

Identify the coordinates: Start at (0, 0), ends at (3, 4)
Apply the formula: $${\text{{Displacement}}} = \sqrt{(3 - 0)^2 + (4 - 0)^2}$$
Calculate: $${\text{{Displacement}}} = \sqrt{9 + 16} = \sqrt{25} = 5 \text{{ meters}}$$

Validation: The displacement is 5 meters, which matches the calculation.

Example 2 (Intermediate)

Problem: A car travels 10 kilometers east, then 15 kilometers south, and finally 5 kilometers west. What is the total distance traveled and the final displacement?

Step-by-Step Solution:

Total distance: $$10 + 15 + 5 = 20 \text{{ kilometers}}$$
Final displacement: Start at (0, 0), moves to (10, 0), then to (10, -15), and finally to (5, -15)
Apply the formula: $${\text{{Displacement}}} = \sqrt{(5 - 0)^2 + (-15 - 0)^2}$$
Calculate: $${\text{{Displacement}}} = \sqrt{25 + 225} = \sqrt{250} \approx 15.81 \text{{ kilometers}}$$

Validation: The total distance is 20 kilometers, and the final displacement is approximately 15.81 kilometers.

4. Problem-Solving Techniques

Visual Strategy: Use diagrams to represent paths and directions clearly.
Error-Proofing: Always check if the calculated displacement makes sense in the context of the problem.
Concept Reinforcement: Practice problems that involve both distance and displacement to understand their differences.