Position, Displacement & Distance

Three closely related but distinct quantities describe where an object is and how far it has moved: position, displacement, and distance.

Concept

Position $x$

A coordinate. In 1D, $x$ is signed: positive on one side of the origin, negative on the other. Position is a vector (here, a 1-component vector).

Displacement $\Delta x$

Change in position over an interval: $\Delta x = x_f - x_i.$ Displacement is also a vector. It can be positive, negative, or zero, and depends only on initial and final positions — not on the path.

Distance $d$

Total path length covered. Always non-negative scalar.

Relations

• $|\Delta x| \le d$, with equality only when motion is in one direction without reversal.
• An object returning to its starting point: $\Delta x = 0$, but $d > 0$.

1D Sign Convention

Choose a positive direction (e.g., east, or "up"). Quantities point along that direction are positive; opposite are negative.

Worked Example

Q: A particle moves from $x = 0$ to $x = 10\,\text{m}$, then back to $x = 4\,\text{m}$. Find the displacement and the distance.

Solution:

• Displacement: $\Delta x = x_f - x_i = 4 - 0 = +4\,\text{m}$.
• Distance: $|10 - 0| + |4 - 10| = 10 + 6 = 16\,\text{m}$.

So the particle covers $16$ m of path but ends up only $4$ m from the start.

Common Confusions

• "Distance and displacement are the same." — They are equal only for straight-line motion with no reversal.
• "Negative displacement means negative distance." — Distance is never negative. A negative sign indicates direction in 1D.
• "If displacement is zero, the object didn't move." — Wrong. A round trip has $\Delta x = 0$ but $d > 0$.

Key Takeaways

• Position is a coordinate; in 1D it has a sign.
• Displacement = final − initial position; a vector; path-independent.
• Distance = path length; a non-negative scalar.
• $|\Delta x| \le d$; equality only when motion is unidirectional.
• Carefully apply sign conventions in 1D problems.