Lecture Notes: Distance vs Displacement

Key Concepts

• Distance:

  • Defined as the length of the path taken between two points.
  • Represented by purple arrows in the diagram.
  • Example: Walking 200 meters north and 350 meters east results in a total distance of 550 meters (200 + 350).

• Displacement:

  • Defined as the shortest distance between the initial and final positions.
  • Represented by the green dotted line in the diagram.
  • Considered as the "shortcut" from start to end point.

Practice Problem

Problem Statement:

• A student walks his dog from point P, 200 meters north, then 350 meters east. The trip takes 400 seconds.
• Question: Calculate the student's displacement and illustrate it on the diagram.

Solution Steps

1. Understand the Diagram:

   • The path creates a right triangle.
   • Utilize the Pythagorean Theorem: ( a^2 + b^2 = c^2 ).

2. Label and Identify Components:

   • Side "a" = 200 meters.
   • Side "b" = 350 meters.
   • Hypotenuse "c" = Displacement.

3. Apply Pythagorean Theorem:

   • ( 200^2 + 350^2 = c^2 )
   • Calculate:
     • ( 200^2 = 40,000 )
     • ( 350^2 = 122,500 )
   • Sum: ( 40,000 + 122,500 = 162,500 )

4. Solve for Displacement (c):

   • Square root of 162,500: ( c = \sqrt{162,500} = 403 ) meters.

Finalizing the Solution

• Write the Displacement:
  • Annotate the diagram with "403 meters" next to the dotted line.
• Add Directionality:
  • Draw an arrowhead to indicate movement direction from start to end.

Summary

• The student's displacement is 403 meters, calculated using the Pythagorean theorem in the context of a right triangle formed by the walking path.