Q.A student walks 3 m due east from a point O and then walks back 3 m due west to O. Explain, with reference to the meanings of the two terms, why the distance travelled and the displacement for this walk are not the same, and state the value of each.

[!TLDR] Distance is the total path length walked (a scalar); displacement is the net change in position (a vector). Here the walker returns to the start, so distance = 6 m but displacement = 0. [!ANSWER] Distance travelled = 6 m; displacement = 0 m.

The student starts at the point O and walks 3 m due east, reaching a point 3 m east of O. She then walks 3 m due west, which brings her back exactly to O.

Distance travelled (path length). This is the total length of the actual path walked, irrespective of direction, so we simply add the two legs of the journey:

Displacement. This is the net change in position, $\Delta x = x_{\text{final}} - x_{\text{initial}}$. Since the student's final position coincides exactly with her initial position (both are O), the displacement is

This example shows clearly that distance and displacement, though both describe "how far" an object moved, capture fundamentally different information: distance (6 m) reflects the full effort/path taken, while displacement (0 m) reflects only the net effect on position. A non-zero distance together with a zero displacement is only possible because the walker reversed direction and retraced part of her path.

[!ANSWER] Distance travelled = 6 m; displacement = 0 m.

Add both legs of the journey (in the same direction, scalar) for distance; subtract final minus initial position (vector, along the line) for displacement.

• Treating distance and displacement as always equal, which is true only for one-directional motion
• Forgetting displacement can be zero even when the object has clearly moved and covered a positive distance
• Not assigning a sign/direction to displacement when it is non-zero