Distance and Displacement

Before any speed or acceleration can be computed, physics needs a bookkeeping device for where a particle is and how far it has moved. That bookkeeping splits into two ideas students routinely merge: distance, the total ground covered, and displacement, the net change of position. The distinction is not pedantry — NEET recycles "distance 8 m, displacement zero" style items almost every alternate year, and JEE Main hides the same idea inside graph-area questions. This page builds both quantities from scratch: frames of reference and sign conventions first, then the master inequality distance ≥ |displacement|, and finally how to read a position–time graph like a story of the motion. The velocity machinery that follows on the average velocity and average speed page sits directly on these foundations.

What are Distance and Displacement? — Complete Theory

1. Frame of reference, origin and position

To state a position you must first choose a reference point (the origin), a direction of positive measurement along the line, and a clock. The trio — origin, axis and clock — is a frame of reference. In one dimension a particle's position is then a single signed number x: +3 m means "3 m on the positive side of the origin", −3 m means "3 m on the negative side". The sign is not decoration; it is the entire direction information in 1D. Choose the axis badly and every later quantity — velocity, acceleration — inherits the confusion, which is why examiners insist you state your sign convention at the top of any solution. NCERT's advice, worth copying into every solution you write: draw the axis, mark the origin, then never silently flip the convention midway.

2. Displacement: the signed change of position

If a particle moves from position x₁ at time t₁ to position x₂ at time t₂, its displacement is

Δx = x₂ − x₁ (a vector along the x-axis; magnitude |Δx|, SI unit m).

Displacement is change of position, not distance walked: it knows nothing about the route, only the endpoints, and it carries the sign of its direction. Walking 8 m east and then 3 m west lands you 5 m east of the start: displacement = +5 m (taking east positive). The same walk described with west positive would give −5 m — same physics, different labels, which is why answers must always quote the convention. Dimensional check: [x₂ − x₁] = [L], consistent, since a difference of two lengths is a length. Displacement can be positive, negative or zero — and each of the three carries real information: positive means net motion along +axis, negative means net motion along −axis, zero means the particle returned home (or never moved).

3. Distance (path length): the scalar odometer

The distance (NCERT's path length) is the total length of the actual path traced, irrespective of direction. It is a scalar: it accumulates, never decreases, and is never negative. For the 8 m east + 3 m west walk the distance is 8 + 3 = 11 m even though the walker ends up only 5 m from home. Distance answers "how much ground was covered"; displacement answers "how far am I from where I began, and in which direction". Both share the SI unit metre and the dimension [L], but they behave differently under reversal — distance keeps adding, displacement starts undoing itself.

4. The master inequality: distance ≥ |displacement|

For any motion, d ≥ |Δx|, with equality if and only if the motion is along a straight line without any reversal of direction. Every reversal adds extra path that displacement ignores, so the gap between the two opens exactly where turning happens. This single sentence powers a whole family of NTA questions: "A body travels from A to B and back…" (d = 2·AB, s = 0), "For which motion are distance and displacement equal?" (no reversal), "Can |displacement| exceed distance?" (never — the shortest possible route between endpoints is the straight line itself). Treat the inequality as a sanity check on every numerical answer you produce in this chapter: if your computed |displacement| ever exceeds your computed distance, you have made a sign or bookkeeping error somewhere.

Visualising Distance and Displacement

The same walk drawn as a position–time graph would rise from 0 to +8 m, then fall to +5 m: rising segments are motion along +x, falling segments motion along −x, and the steeper the segment, the faster the motion. That graph-reading skill is developed properly on the instantaneous velocity page; for now, note that the net rise or fall of the graph equals the displacement, while the total up-and-down travel of the pen equals the distance. Keep the two images — track and graph — paired in your head; NTA questions flip between them freely.

Solved Examples on Distance and Displacement (Step-by-Step)

Solved Example 1 — Round trip: distance 18 m, displacement 0

Q. A particle's position (in metres) along the x-axis is x = 6t − t². Find the distance travelled and the displacement for the interval t = 0 to t = 6 s.

Step 1. Endpoints: x(0) = 0, x(6) = 36 − 36 = 0 → displacement = 0 − 0 = 0.

Step 2. But the particle need not have stayed put. v = dx/dt = 6 − 2t = 0 at t = 3 s — the particle reverses there (a full treatment is on the acceleration page).

Step 3. Turning point position: x(3) = 18 − 9 = 9 m. Outward leg: 0 → 9 m (9 m); return leg: 9 m → 0 (9 m).

Step 4. Distance = 9 + 9 = 18 m; displacement = 0. The inequality d ≥ |s| reads 18 ≥ 0 ✔.

Solved Example 2 — Sign convention does the work

Q. A bus on a straight highway moves 300 m north, then 400 m south. Taking north as positive, find distance and displacement; then repeat with south as positive.

Step 1. Distance is convention-independent: 300 + 400 = 700 m (a scalar never changes with axis choice).

Step 2. North positive: +300 + (−400) = −100 m → 100 m south.

Step 3. South positive: −300 + 400 = +100 m → 100 m south. Same physics; only the label flips.

Practice Questions on Distance and Displacement (With Solutions)

Practice Q1 (Numerical-value type)

Q. A man walks 5 m towards east, then 12 m towards north. Taking the straight-line separation, the magnitude of his displacement is ______ m. (Hint: this two-leg check still uses the inequality — verify d = 17 m > 13 m.)

Solution. The legs are perpendicular, so the net shift is the vector sum: √(5² + 12²) = √169 = 13 m. Distance is 5 + 12 = 17 m, and indeed 17 ≥ 13 ✔.

Practice Q2 (Single-correct MCQ)

Q. In which case is distance equal to the magnitude of displacement? (a) a car rounds a circular track once (b) a lift goes up 10 m and comes down 10 m (c) a stone rolls 20 m down a straight incline without turning (d) a bee flies to a flower 5 m away and returns

Solution. Equality demands straight-line motion with no reversal. Options (a), (b), (d) all return to the start (s = 0, d ≠ 0). Only (c) is reversal-free.

Practice Q3 (Assertion–Reason)

Q. Assertion (A): The displacement of a particle that completes one full lap of a closed circular track is zero. Reason (R): Displacement depends only on the initial and final positions, not on the path taken.
Options: (a) both true, R explains A · (b) both true, R does not explain A · (c) A true, R false · (d) A false, R true

Solution. A closed lap returns the particle to its start, so x₂ = x₁ and Δx = 0 — precisely because displacement is endpoint-defined. R is the governing principle and explains A.

Key Formulas & Takeaways

FAQs on Distance and Displacement

What is the difference between distance and displacement?

Distance is the total path length travelled — a scalar that never decreases. Displacement is the straight-line change in position x₂ − x₁ — a vector that can be positive, negative or zero.

Can displacement be greater than distance?

Never. The magnitude of displacement is the shortest separation between the two endpoints, so distance ≥ |displacement| always, with equality only in straight-line motion without reversal.

Can displacement be zero while distance is not?

Yes — any motion that returns to its starting point has zero displacement but non-zero distance; walking 4 m east and then 4 m west gives 8 m of distance and zero displacement.

How is position read on a position-time graph?

The height of the curve at each time is the position coordinate; a rising curve means motion in the positive direction, a falling curve motion in the negative direction, and a horizontal line means rest.