Position, Displacement & Distance

A position answers four questions before it answers “where”: measured from what, in which direction, in what unit, and at what time. Put the origin at a school gate, point +x east, measure in metres and start the clock when you do, and a parked car 35 m east of the gate is at x = +35 m. Put the origin at a bus stop 50 m east of the gate instead and the same car is at x = −15 m. Neither description is more correct. The sign says only which side of the origin the car is on; a negative position is not “less” of anything, and it says nothing about motion. The clock is a choice too: t = 0 can be the moment you press start, not the moment the object began to move. Write the frame down, in words or on a sketch, because every sign that follows inherits it.

Over an interval [t₁, t₂], with t₂ after t₁, the displacement is Δx = x(t₂) − x(t₁). The shorthand x₁ means x(t₁), the position at the instant t₁. Shift the origin to x₀ and every position becomes x − x₀, so Δx′ = (x₂ − x₀) − (x₁ − x₀) = Δx: the choice of zero drops out of every displacement. Reverse the axis and every position changes sign, so Δx does too — the motion is the same, only its description flips. A cart that starts at the 2.0 m mark of a track and ends at the 7.0 m mark has Δx = +5.0 m measured from the track's zero mark, or from a post at the 5.0 m mark, where it runs from −3.0 m to +2.0 m. With +x pointing the other way it is −5.0 m in both. Its final position equals its displacement only in the frame whose origin sits at the start.

Distance is path length, so it depends on everything the object did inside the interval. The recipe: find each instant τ where the motion reverses, cut [t₁, t₂] there, take |Δx| for every piece, and add. Reversals are local maxima and minima of x(t), so with a formula in hand you find them where dx/dt = 0 and changes sign — the derivative the next lesson names velocity. For x(t) = 2 + 6t − t² in metres, dx/dt = 6 − 2t vanishes at τ = 3 s, where x = 11 m. Over [0, 5 s] the cart runs 2 m → 11 m → 7 m, so d = 9 m + 4 m = 13 m while Δx = +5 m. Extra cuts at points that are not reversals, such as where x crosses zero or at round clock times, do no harm and no work; a missed reversal always undercounts. In integral form the same recipe reads d = ∫ |dx/dt| dt, taken from t₁ to t₂.

Each piece of path adds its full length to d but only its signed length to Δx, and signed lengths can cancel: |a + b| ≤ |a| + |b|. So d ≥ |Δx| always, with equality exactly when x(t) never reverses on the interval. This matters most when positions come from a sensor rather than a formula. Exact readings at both endpoints fix the displacement completely, however sparse the intermediate readings. Adding |Δx| between successive exact readings gives a lower bound on distance over the same interval. It is exact if no gap hides a reversal; noisy readings need separate treatment and can overestimate the distance. Sample the cart with x(t) = 2 + 6t − t² every 2 s over [0, 6 s] and the readings 2, 10, 10 and 2 m give 8 + 0 + 8 = 16 m, while the cart really covered 18 m: the run from 10 m out to 11 m and back, inside the gap [2 s, 4 s], left no trace. Sample only at 0 and 6 s and the estimate collapses to zero.

Displacements over adjacent intervals add with their signs: Δx over [t₁, t₃] equals Δx over [t₁, t₂] plus Δx over [t₂, t₃], because the middle position x(t₂) cancels. Distances add too, as lengths. Magnitudes of displacement do not. For x(t) = 2 + 6t − t², Δx over [0, 3 s] is +9 m and over [3 s, 6 s] is −9 m, so over [0, 6 s] Δx = 0 — the cart is back where it started — while d = 9 m + 9 m = 18 m. That is why a round trip is the cleanest test of the two ideas: zero displacement, full distance. Keep the interval attached to every symbol, written out as “Δx over [0, 3 s]” or compactly as Δx₁₂ once t₁ and t₂ are defined. A Δx with no interval named is a number without a question, and most sign errors in later kinematics start there.

Three assumptions sit under everything above. First, the object is a point: a train 200 m long has no single position until you name the point on it you are tracking, and its front and back pass a signal at different times. Second, the motion runs along one line. A coordinate s measured along a curving road still gives distance, as |Δs| when the car never reverses, but Δs is not the straight-line displacement: one lap of a 400 m track is 400 m of path and zero displacement, because start and finish coincide. Third, the origin must stay put. Every origin at rest on the track gives the same Δx, but a passenger who walks 12 m toward the back of a train while the train rolls 30 m forward has Δx = −12 m in the train's frame and +18 m in the ground's, taking +x forward in both. Converting between moving frames belongs to relative motion; here it is enough to know the rule has that edge.