In short

Position is where you are on a number line, measured from a reference point (the origin). Distance is the total length of the path you travel — always zero or positive. Displacement is the change in position — a signed quantity that tells you how far and in which direction you ended up from where you started. Distance and displacement are equal only when you move in one direction without turning back.

You step out of your house and walk 300 m east to the neighbourhood chai stall. You buy a cup of cutting chai, realise you forgot your wallet, and walk 400 m west back past your house to borrow money from a friend. How far did you walk? And how far are you from home?

These are two different questions. The first asks about distance — the total ground your feet covered. The second asks about displacement — the gap between where you started and where you ended up, measured in a straight line with a direction. The answer to the first question is 700 m. The answer to the second is 100 m west of home. Same walk, two different numbers, because the two quantities measure fundamentally different things.

This distinction — between "how much path did you cover" and "how far did you actually get" — is the foundation of kinematics. Every formula you will meet in motion, from speed to velocity to acceleration, depends on whether you are talking about distance or displacement. Get them confused, and every answer that follows will be wrong.

Position — a coordinate on a number line

Before you can talk about how far something moved, you need to say where it is. That requires two choices: a reference point (called the origin) and a direction to call positive.

Think of a straight road through a small town. Pick the bus stop as your origin. Everything to the east gets a positive coordinate; everything to the west gets a negative coordinate. Your house is 200 m east of the bus stop, so your position is +200 m. The school is 150 m west of the bus stop, so the school's position is -150 m.

Number line showing position as a coordinate A horizontal number line with the bus stop at the origin (0). The school is at minus 150 m to the left. The house is at plus 200 m to the right. Arrows show the positive direction pointing east. +x (east) 0 bus stop −150 m school +200 m house
Position is a number on a line, measured from a chosen origin. The sign tells you the direction.

The position of an object is written as x (or \vec{r} in two or three dimensions). It is not a property of the object alone — it depends on your choice of origin. Move the origin from the bus stop to the school, and the house's position changes from +200 m to +350 m. The house did not move; your coordinate system did.

This is the first lesson: position is always relative to a reference point. There is no "absolute position" in physics. When someone says "the train is at kilometre 342," they mean 342 km from the starting station — that station is the origin.

Distance — total path length

Distance answers the question: how many metres did the object actually travel?

If you walk 300 m east and then 400 m west, you walked a total of 300 + 400 = 700 m. It does not matter that part of the walk cancelled out the other part. Your legs carried you 700 m. Your shoes wore out by 700 m worth of friction. The odometer on an autorickshaw does the same thing — it adds up every metre of road the vehicle covers, regardless of direction.

Distance is a scalar — it has magnitude but no direction. It is always zero or positive. You cannot walk a negative distance. Even if you pace back and forth in your room all night, your distance is the sum of every back and every forth.

Mathematically, if an object moves from position x_1 to x_2, then back to x_3, the total distance is:

d = |x_2 - x_1| + |x_3 - x_2|

Why the absolute values: each segment of the path contributes its full length, regardless of whether the object moved in the positive or negative direction. The vertical bars ensure every contribution is positive.

Displacement — change in position

Displacement answers a different question: how far is the object from where it started, and in which direction?

If you walk 300 m east and then 400 m west, your displacement is the difference between your final position and your initial position. You started at home (call it x_i = 0). You ended at 100 m west of home (x_f = -100 m). So your displacement is:

\Delta x = x_f - x_i = -100 - 0 = -100 \text{ m}

Why: the Greek letter \Delta (delta) always means "change in" — final value minus initial value. The negative sign tells you the displacement is in the negative direction (west).

Displacement is a vector — it has both magnitude and direction. In one dimension, the direction is encoded in the sign: positive means east (or right, or forward), negative means west (or left, or backward). In two or three dimensions, displacement is a vector with components.

The magnitude of the displacement, |\Delta x| = 100 m, is how far you ended up from home. The sign tells you which side.

Path versus displacement for back-and-forth motion A person starts at home, walks 300 m east to the chai stall, then 400 m west to a friend's house. The total path (distance) is 700 m, shown as a curved dotted line. The displacement is 100 m west, shown as a straight red arrow from home to the final position. east home x = 0 chai stall x = +300 m friend x = −100 m 300 m east → ← 400 m west displacement = 100 m west
The dashed curves show the actual path (total: 700 m). The solid red arrow shows the displacement (100 m west). Distance counts every step; displacement only cares about start and finish.

When distance and displacement differ — back-and-forth motion

Distance and displacement are equal when you move in a straight line without turning around. Walk 500 m east without stopping: distance = 500 m, displacement = +500 m. They agree.

The moment you turn around, they diverge. Every metre you retrace adds to your distance but subtracts from your displacement. This is why:

The general rule:

\text{displacement magnitude} \leq \text{distance}

Why: the straight line between two points is always the shortest path. Any detour adds to the distance without changing the displacement. Equality holds only when the path itself is a straight line in one direction.

The table below makes this concrete for several everyday situations:

Situation Distance Displacement
Walk 500 m east 500 m +500 m (east)
Walk 500 m east, then 500 m west 1,000 m 0
Walk 500 m east, then 200 m west 700 m +300 m (east)
One full lap around a 400 m track 400 m 0
Earth completes one full orbit around the Sun ~940 million km 0

Notice the last row. The Earth travels nearly a billion kilometres every year, but its displacement after one complete orbit is zero — it ends up right where it started.

The position–time graph and what its shape tells you

A position–time graph plots the object's position x on the vertical axis against time t on the horizontal axis. Every point on the graph tells you where the object is at that instant. The shape of the graph tells you how the object is moving.

Here is a position–time graph for a student who walks east for 3 seconds, pauses for 2 seconds, and then walks west for 3 seconds:

A position–time graph for back-and-forth motion. The rising segment means the student moves east (position increasing). The flat segment means they are stationary. The falling segment means they move west (position decreasing).

Here is how to read this graph:

The position–time graph is a complete record of the motion. From it, you can extract both distance and displacement over any time interval — but the methods are different.

Reading displacement from the graph

Displacement is the change in position: \Delta x = x_f - x_i. On the graph, this is the vertical difference between the final and initial points.

From t = 0 to t = 8 s in the graph above:

\Delta x = x(8) - x(0) = -3 - 0 = -3 \text{ m}

The displacement is -3 m — the student ended up 3 m in the negative direction from where they started.

Reading distance from the graph

Distance is the total path length — you add up every segment's length regardless of direction. You cannot just read the vertical difference; you must account for every change in direction.

From t = 0 to t = 8 s:

d = 6 + 0 + 9 = 15 \text{ m}

Why the segments are added separately: each time the object changes direction, you have a new segment. Distance accumulates the absolute length of every segment. Displacement only looks at start and end.

The student's distance (15 m) is five times their displacement magnitude (3 m). The back-and-forth motion inflated the distance far beyond the net change in position.

Worked examples

Example 1: Walking east then west — number line

A student leaves her house and walks 300 m east to a bookshop. She then walks 400 m west to meet a friend. Find (a) the total distance she walks, and (b) her displacement from her house.

Number line for Example 1: walking east then west A number line with the house at the origin (0 m). The bookshop is at plus 300 m to the east. The friend's house is at minus 100 m to the west. An arc shows the 300 m eastward walk, and another arc shows the 400 m westward return. The displacement arrow points from the origin to minus 100 m. east 0 house +300 m bookshop −100 m friend 300 m → ← 400 m Δx = −100 m
The dashed arcs show the two legs of the walk. The solid red arrow shows the net displacement.

Step 1. Set up a coordinate system.

Choose the house as the origin. East is the positive direction.

  • Initial position: x_i = 0
  • After the eastward walk: x_1 = +300 m
  • After the westward walk: x_f = 300 - 400 = -100 m

Why: walking west means subtracting from the position. 300 m east followed by 400 m west puts you 100 m west of the start, which is -100 m.

Step 2. Calculate the total distance.

d = 300 + 400 = 700 \text{ m}

Why: distance is the sum of all path lengths, regardless of direction. Each leg contributes its full length.

Step 3. Calculate the displacement.

\Delta x = x_f - x_i = -100 - 0 = -100 \text{ m}

Why: displacement is final position minus initial position. The negative sign means the student ended up west of her starting point.

Result: Distance = 700 m. Displacement = -100 m (100 m west of home).

What this shows: The student walked 700 m total but ended up only 100 m from home. The back-and-forth motion made the distance seven times larger than the displacement magnitude. If you asked "how far did she walk?" and "how far is she from home?" you would get very different answers — and both would be correct, because they measure different things.

Example 2: Reading a position–time graph

The position–time graph below shows the motion of an autorickshaw along a straight road over 10 seconds. Read the graph to find (a) the displacement and (b) the total distance travelled between t = 0 and t = 10 s.

Position–time graph for the autorickshaw. The line rises from A to B (moving forward), stays flat from B to C (stopped), then falls from C to D (moving backward). Read the coordinates of each labelled point to find distance and displacement.

Step 1. Read the key coordinates from the graph.

Point Time (s) Position (m)
A 0 0
B 4 20
C 6 20
D 10 0

Why: these four points mark every change in the autorickshaw's motion — it moves forward (A to B), stops (B to C), then reverses (C to D). Each transition is a new segment for the distance calculation.

Step 2. Calculate the displacement from A to D.

\Delta x = x_D - x_A = 0 - 0 = 0

Why: the autorickshaw starts at x = 0 and ends at x = 0. It returned to its starting position. The displacement is zero even though the auto moved a lot in between.

Step 3. Calculate the total distance from A to D.

Break the motion into segments at every turning point:

  • A to B: |20 - 0| = 20 m
  • B to C: |20 - 20| = 0 m (stationary)
  • C to D: |0 - 20| = 20 m
d = 20 + 0 + 20 = 40 \text{ m}

Why: the auto first moves 20 m forward and then 20 m backward. Both legs contribute their full length to the distance. The absolute values ensure you are adding positive lengths even when the position decreases.

Result: Displacement = 0 m. Distance = 40 m.

What this shows: The autorickshaw travelled 40 m total but ended up exactly where it started. This is the round-trip scenario — maximum disagreement between distance and displacement. The position–time graph makes this visually obvious: the graph starts and ends at the same height (same position), but the path it traces covers real vertical distance in between.

Common confusions

If you understand what position, distance, and displacement mean, how to compute them, and how to read them from a position–time graph, you have the tools you need for the next topics in kinematics. What follows is for readers who want the formal vector treatment and a deeper look at signed displacement in one dimension.

Displacement as a vector in two and three dimensions

In one dimension, displacement is a signed number: \Delta x = x_f - x_i. The sign carries the direction.

In two dimensions, position is a vector \vec{r} = x\,\hat{i} + y\,\hat{j}. Displacement becomes:

\Delta\vec{r} = \vec{r}_f - \vec{r}_i = (x_f - x_i)\,\hat{i} + (y_f - y_i)\,\hat{j}

Why: displacement is still final minus initial, but now it is a vector subtraction. Each component is an independent signed displacement along that axis.

The magnitude of the displacement vector is:

|\Delta\vec{r}| = \sqrt{(x_f - x_i)^2 + (y_f - y_i)^2}

Why: this is the Pythagorean theorem applied to the displacement components. The magnitude gives you the straight-line distance between the starting and ending positions.

For example, if you walk 300 m east and then 400 m north, your displacement vector is \Delta\vec{r} = 300\,\hat{i} + 400\,\hat{j} m, and its magnitude is \sqrt{300^2 + 400^2} = \sqrt{250000} = 500 m. Meanwhile, the distance you walked is 300 + 400 = 700 m — strictly greater than the displacement magnitude, as expected.

Why displacement doesn't depend on the choice of origin

Suppose you shift your origin by a constant a — so every position x becomes x' = x - a. The displacement in the new coordinate system is:

\Delta x' = x_f' - x_i' = (x_f - a) - (x_i - a) = x_f - x_i = \Delta x

Why: the constant shift a cancels out when you take the difference. This is why displacement is a more fundamental quantity than position in physics — it is independent of where you put the origin.

This is not a coincidence. Displacement is a vector (a directed change), while position is a point (a location relative to an origin). In physics, the laws of motion involve changes in position (velocity, acceleration), not positions themselves. That is why the equations of motion work regardless of where you place your origin.

Signed area under a velocity–time graph

You will see in the next topic (Speed and Velocity) that velocity is the rate of change of displacement. This leads to an important connection: the displacement over a time interval equals the signed area under the velocity–time graph over that interval, while the distance equals the total (unsigned) area. When velocity is negative (motion in the negative direction), the signed area subtracts from the displacement but the unsigned area still adds to the distance. This is the graphical version of everything you have learned in this article.

Where this leads next