Displacement Physics Example 4

Follow the full solution, then compare it with the other examples linked below.

Example 4

medium
A hiker walks 3 km3 \text{ km} north, then 4 km4 \text{ km} east. What is the hiker's total distance walked and their displacement (magnitude and direction)?

Solution

  1. 1
    Total distance = 3+4=7 km3 + 4 = 7 \text{ km} (scalar — path length).
  2. 2
    Displacement magnitude: d=32+42=9+16=25=5 kmd = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5 \text{ km}.
  3. 3
    Direction: θ=tan⁡−1(43)≈53.1°\theta = \tan^{-1}\left(\frac{4}{3}\right) \approx 53.1° east of north.

Answer

Distance=7 km;Displacement=5 km at 53.1° east of north\text{Distance} = 7 \text{ km}; \quad \text{Displacement} = 5 \text{ km at } 53.1° \text{ east of north}
Distance is the total path length (always positive). Displacement is the straight-line vector from start to finish. This classic 3-4-5 triangle shows how displacement can be much shorter than distance.

About Displacement

The change in position of an object, measured as the straight-line distance and direction from the starting point to the ending point.

Learn more about Displacement →

More Displacement Examples

Example 1 easy

A person walks [formula] east and then [formula] north. What is the total distance traveled and the

Example 2 medium

A runner completes one full lap around a [formula] circular track. What is the runner's displacement

Example 3 easy

A car drives [formula] north and then [formula] south. What is the displacement?

← All Displacement Examples Displacement Concept

Related Concepts

PositionVelocityVectors