Distance vs Displacement: The Difference

Distance is the length of the path travelled, a scalar with no direction. Displacement is the change in position, a vector drawn straight from start to finish. Distance depends on the route, displacement only on the endpoints, and for a round trip the displacement is exactly zero.

AP Physics: Unit 1 (topics 1.1 Scalars and Vectors in One Dimension, 1.2 Displacement, Velocity, and Acceleration, 1.3 Representing Motion). AP Physics 1 Unit 1, Kinematics, carries 10 to 15 percent of the multiple-choice section over an estimated 12 to 17 class periods. EK 1.2.A.2 defines displacement as the change in an object's position, with the relevant equation delta x = x - x_0. EK 1.1.A.3 names distance as an example of a scalar quantity and displacement as an example of a vector quantity; the CED gives no separate definition of distance. EK 1.1.A.3.ii states that vector notation is not required for components along an axis and that in one dimension the sign of the component completely describes its direction, and EK 1.1.B.1 that opposite directions are denoted by opposite signs. EK 1.3.A.4.iii makes the displacement equal to the area bounded by a velocity against time curve and the horizontal axis. The Unit 3 material quoted here is EK 3.2.A.3 and 3.2.A.3.i, which describe work in terms of the displacement of the point of application, and EK 3.2.A.4.iii, which equates the energy dissipated by friction to the friction force times the length of the path; EK 3.2.A.1.ii and 3.2.A.1.iv contrast path-independent conservative work with path-dependent nonconservative work. Suggested skills are 1.A, 2.C, 3.B and 3.C for Topic 1.1; 1.C, 2.B, 2.C and 3.C for Topic 1.2; and 1.C, 2.A, 2.C and 3.B for Topic 1.3.

The distinction, stated once

Distance answers how far you went. Displacement answers how far you ended up from where you started, and in which direction.

The AP Physics 1 CED defines only one of the two. EK 1.2.A.2 says displacement is the change in an object's position, with the relevant equation $\Delta x = x - x_0$ . Distance is not given a definition anywhere in the course framework; EK 1.1.A.3 simply names it, listing distance and speed as examples of scalar quantities while position, displacement, velocity, and acceleration are examples of vector quantities.

That asymmetry is not an oversight. Displacement needs a definition because it is built from two positions and a subtraction, and getting the order and the signs right is the whole skill. Distance needs no definition because it is just the length of the path, which is what a tape measure laid along the route would read.

The consequence that matters: displacement forgets the route entirely. Give it the start point and the end point and it is fully determined, whatever happened in between. Distance forgets nothing; every metre travelled is counted, in both directions, so it can only grow.

The two carry the same unit, the metre, which is why they get swapped. And the two are the source of the speed vs velocity distinction one level up: divide a distance by a time and you get an average speed, divide a displacement by a time and you get an average velocity. If those two averages ever disagree in a problem, this page is where the disagreement came from. Each side on its own sits in the displacement glossary entry.

Distance vs displacement, side by side

	Distance	Displacement
What it is	The length of the path actually travelled	The change in position, from start point to end point
CED source	Named as a scalar in EK 1.1.A.3	Defined in EK 1.2.A.2, $\Delta x = x - x_0$
Scalar or vector	Scalar	Vector
Symbol	$d$	$\Delta \vec{x}$ , or $\Delta x$ for a component along an axis
SI unit	metre	metre, the same unit
Can be negative	No	Yes in one dimension, where the sign is the direction
Depends on the route taken	Yes, entirely	No, only on the two endpoints
Can decrease as the motion continues	No, it only accumulates	Yes, if the object turns back toward its start
Value for a complete round trip	The full path length	Exactly zero
Combining two legs in two dimensions	Add the lengths	Add as vectors, magnitude from Pythagoras
Which is never the smaller of the two	Distance, always
From a velocity against time graph	Total area, all of it counted positive	Signed area, from EK 1.3.A.4.iii
Divide by elapsed time and you get	Average speed	Average velocity, from EK 1.2.B.2
What a car odometer reads	This	Not this
On the AP Physics 1 equation sheet	The symbol key reads $d$ = distance	The symbol key reads $x$ = position; the displacement appears as $(x - x_0)$ inside a kinematic equation

Read the last two rows together and the practical rule falls out. The sheet gives the two quantities different letters, and it means it. A $d$ in a printed equation is not automatically interchangeable with an $x - x_0$ , and the section below on the work equation is where that bites.

The case that separates them: turn a corner

The cleanest separation needs only two legs at right angles, no reversing at all.

Walk 6.0 m east, then 8.0 m north. The distance travelled is $6.0 + 8.0 = 14.0\ \text{m}$ , because distance just adds. The displacement is not 14.0 m, because the two legs point in different directions and vectors do not add like that. The magnitude is the hypotenuse:

\lvert \Delta \vec{x} \rvert = \sqrt{(6.0)^2 + (8.0)^2} = \sqrt{100} = 10.0\ \text{m}

at $53^\circ$ north of east. Four metres of the walking simply does not show up in the displacement, and no amount of care with signs would recover it, because nothing went wrong.

Now add a reversal and the gap widens further:

Journey	Distance	Displacement	Ratio
10 m east	10 m	10 m east	1.00
6 m east then 8 m north	14 m	10 m at $53^\circ$ north of east	0.71
6 m east then 2 m west	8 m	4 m east	0.50
5 m east then 5 m west	10 m	0	0.00
One lap of a 400 m track	400 m	0	0.00

The first row is the case where the two agree, and it is the only shape of journey for which they do: a straight line, travelled in one direction. Every other row loses something, and what it loses is exactly the part of the path that did not move the object further from its start.

The inequality this guarantees. For any journey at all,

d \geq \lvert \Delta \vec{x} \rvert

because the straight line between two points is the shortest path between them. If a calculation ever hands you a distance smaller than the magnitude of the displacement, the error is arithmetic and this inequality has caught it.

One small convenience about that $53^\circ$ : the AP Physics 1 equation sheet's table of trigonometric values for common angles includes both $37^\circ$ and $53^\circ$ , with $\sin 53^\circ = 4/5$ and $\cos 53^\circ = 3/5$ . Those are the two non-right angles of a 3-4-5 triangle, which is why so many exam displacements come out to clean numbers.

In one dimension the displacement is a signed number

Along a single axis you do not need arrows, and the CED says so explicitly. EK 1.1.A.3.ii states that vector notation is not required for vector components along an axis, and that in one dimension the sign of the component completely describes the direction of that component. EK 1.1.B.1 adds that when determining a vector sum in a given one-dimensional coordinate system, opposite directions are denoted by opposite signs.

So a one-dimensional displacement is a number with a sign, and the sign is doing the work an arrow would do in two dimensions. Two things follow, and both are places marks go.

First, you must declare the axis before you write any number. "The displacement is $-5\ \text{m}$ " is meaningless until the reader knows which way is positive. Declare it once, then hold it for the entire problem. An axis that flips halfway through a solution produces signs that are individually defensible and collectively wrong.

Second, the subtraction order is fixed. $\Delta x = x - x_0$ means final position minus initial position, never the other way round. An object that moves from $x_0 = 9\ \text{m}$ to $x = 4\ \text{m}$ has $\Delta x = 4 - 9 = -5\ \text{m}$ , a displacement of 5 m in the negative direction, while the distance travelled is $5\ \text{m}$ with no sign at all. Reverse the subtraction and you get the right magnitude with the wrong direction, which in a kinematics problem then flips the sign of a velocity and an acceleration downstream of it.

A useful check: the sign of the displacement and the sign of the average velocity always match, because the elapsed time in $\vec{v}_{\text{avg}} = \Delta \vec{x}/\Delta t$ is positive. If your displacement is negative and your average velocity came out positive, one of them is wrong.

Reading each one off a graph

This is where the difference becomes a procedure rather than a definition, and it is the form the exam uses most often.

On a graph of velocity against time, EK 1.3.A.4.iii says the displacement of an object during a time interval is equal to the area under the curve, described as the area bounded by the function and the horizontal axis for the appropriate interval. The word doing the work is "bounded": area above the axis counts positive, area below counts negative, and you add them with their signs.

So from one graph you can read both quantities, and the two procedures differ by exactly one step:

Displacement: compute each area, attach a sign according to which side of the axis it is on, and add.
Distance: compute each area, take its absolute value, and add.

That is the entire difference. Worked example two below runs both on the same graph and gets $+2\ \text{m}$ and $22\ \text{m}$ .

On a graph of position against time, the displacement over an interval is the change in the vertical coordinate, $x_{\text{final}} - x_{\text{initial}}$ , read straight off the axis. The distance is not readable that simply: you have to find every turning point, where the graph changes from rising to falling or back, and add the size of each rise and fall separately. A position graph that goes up to 8 m, back down to 3 m, and up again to 6 m has a displacement of $+6\ \text{m}$ and a distance of $8 + 5 + 3 = 16\ \text{m}$ .

A turning point on a position graph is where a distance calculation gains a term and a displacement calculation does not. If a graph has no turning points, the two agree, which is the graphical version of "never reversed direction".

The same letter d, two different meanings

Here is a place where this distinction is genuinely subtle rather than merely careless, and it sits in Unit 3 rather than Unit 1.

The AP Physics 1 equation sheet prints the work done by a constant force as

W = F_{\parallel} d = F d \cos\theta

and its symbol key reads $d$ = distance. But the CED prose that introduces that equation does not say distance. EK 3.2.A.3 says the amount of work done on a system by a constant force is related to the components of that force and the displacement of the point at which that force is exerted, and EK 3.2.A.3.i says only the component of the force parallel to the displacement of the point of application will change the system's total energy.

Then a few lines later the CED uses the same letter the other way. EK 3.2.A.4.iii says the energy dissipated by friction is typically equated to the force of friction times the length of the path over which the force is exerted, and writes it as $\Delta E_{\text{mech}} = F_f d \cos\theta$ .

So within one topic the letter $d$ carries a displacement in one relation and a path length in another, and the difference is not cosmetic:

Work by a constant force uses the displacement of the point of application. Push a crate 4 m out and 4 m back and the work you did is not zero, because you must apply the equation to each leg with its own displacement and its own angle. Doing it in one step with the net displacement of zero gives zero work, which is wrong.
Energy dissipated by friction uses the path length directly. Over the same out-and-back trip, friction dissipates energy proportional to the full 8 m, and there is no cancellation available.

The underlying reason is in EK 3.2.A.1: the work done by a conservative force is path-independent and depends only on the initial and final configurations, while EK 3.2.A.1.iv says the work done by a nonconservative force is path-dependent, and EK 3.2.A.1.v names friction and air resistance as examples. Conservative forces care about displacement. Nonconservative forces care about the path. That is the distance-and-displacement distinction promoted to a statement about energy, conservative vs nonconservative force takes it further, and worked example three puts numbers on it.

Where it costs a mark

Adding leg lengths to get a displacement. In two dimensions you must add the legs as vectors. Six metres east plus eight metres north is a displacement of ten metres, not fourteen.
Reporting a displacement with no direction and no declared axis. Displacement is a vector. A sign is sufficient in one dimension, but only once the reader knows which way is positive.
Reversing the subtraction. $\Delta x = x - x_0$ , final minus initial. The other order gives a displacement pointing the wrong way, and everything computed from it inherits the error.
Reporting a negative distance. Distance is a path length and cannot be negative. If your working produced a minus sign, you computed a displacement.
Summing unsigned areas under a velocity graph and calling the result a displacement. Unsigned areas give the distance. EK 1.3.A.4.iii wants the area bounded by the function and the axis, which means area below the axis subtracts.
Using the net displacement in $W = Fd\cos\theta$ across a journey that reversed. Apply the work equation leg by leg. The net displacement of a round trip is zero and the work done by a push around it is not.
Using the displacement instead of the path length for friction. EK 3.2.A.4.iii is explicit that the dissipated energy goes with the length of the path.
Substituting a path length into $v_x^2 = v_{x0}^2 + 2 a_x (x - x_0)$ . That bracket is a displacement along the axis. A projectile that goes up and comes back down has travelled a nonzero path and has $x - x_0 = 0$ in the vertical direction, and the equation will honour whichever number you give it.
Assuming the displacement is the shortest road, not the shortest line. The displacement is the straight vector between the endpoints, even if no route along it exists. That is a feature: it is what makes displacement independent of the route.

When they coincide, and what the equation sheet gives you

The two quantities are the same number under one condition: the motion is along a straight line and never reverses direction. Then the path lies along the displacement, the two measurements coincide, and the only remaining difference is that one of them carries a direction.

That condition holds for nearly every problem in a first pass through kinematics, which is exactly why the distinction goes unnoticed. A cart accelerating down a ramp. A dropped ball, until it hits. A car braking in a straight line to a stop. In all of those you can use the words interchangeably and never be caught.

What breaks it, in the order it usually turns up: a projectile, because the path is a curve and the displacement is the chord; anything thrown up and caught, because the vertical displacement returns to zero; circular motion, where a full revolution has a displacement of zero; and any energy problem involving friction, because the dissipation follows the path while the potential-energy change follows the endpoints.

On the equation sheet, neither quantity gets a defining line. There is no $\Delta x = x - x_0$ printed and no path-length formula. What the sheet gives you is the letters and the places they appear:

The symbol key distinguishes them: $d$ = distance and $x$ = position, with $y$ = vertical position.
The displacement appears as the bracket $(x - x_0)$ in $v_x^2 = v_{x0}^2 + 2 a_x (x - x_0)$ , and as $\Delta \vec{x}$ in the spring force $\vec{F}_s = -k \Delta \vec{x}$ , where it is a stretch from equilibrium.
The distance appears as $d$ in $W = F_{\parallel} d = F d \cos\theta$ and in $\Delta K = \sum W_i = \sum F_{\parallel, i}\, d_i$ .
$\Delta U_g = m g \Delta y$ uses a vertical displacement, and the $\Delta$ is the signal: a rise is positive, a fall is negative, and a round trip gives zero.

The full sheet is here. For the CED framing, Topic 1.1 covers the scalar and vector split, Topic 1.2 defines the displacement, and Topic 1.3 covers the graph areas. The procedures that use these live in kinematic equations and, for the two-dimensional case, how to solve projectile motion problems.

Two legs at right angles, then a reversal

A hiker walks 6.0 m east, then 8.0 m north, then 3.0 m south. Find the total distance travelled and the displacement, giving the displacement as a magnitude and a direction. Take east and north as the positive directions.

Declare the axes: east is positive $x$ , north is positive $y$ . The southward leg is therefore negative $y$ .
Distance. Distance is a scalar and takes no notice of direction, so add the three path lengths: $6.0 + 8.0 + 3.0 = 17.0\ \text{m}$ .
Displacement, component by component. East component: $\Delta x = +6.0\ \text{m}$ , since only the first leg moved the hiker east. North component: $\Delta y = +8.0 - 3.0 = +5.0\ \text{m}$ , the signed sum of the two vertical legs.
Magnitude: $\lvert \Delta \vec{x} \rvert = \sqrt{(6.0)^2 + (5.0)^2} = \sqrt{36 + 25} = \sqrt{61} = 7.81\ \text{m}$ .
Direction: $\theta = \arctan\!\left(\frac{5.0}{6.0}\right) = 39.8^\circ$ north of east. State the reference direction, or the angle means nothing.
Check the inequality. $17.0\ \text{m} \geq 7.81\ \text{m}$ , as it must be. About 9.2 m of walking did not increase the hiker's separation from the start, split between the perpendicular leg and the reversal.
Now compare with a hiker who walks the same 17.0 m in a straight line east. Distance $17.0\ \text{m}$ , displacement $17.0\ \text{m}$ east, and the two are equal. Same distance, very different displacement, which is the whole point: distance cannot tell you where an object ended up.

Distance 17.0 m. Displacement 7.81 m at 39.8 degrees north of east, with components $+6.0\ \text{m}$ east and $+5.0\ \text{m}$ north. A hiker covering the same 17.0 m in one straight line would have a displacement of 17.0 m, so the distance travelled fixes neither the magnitude nor the direction of the displacement.

Both quantities from one velocity graph

An object moves along the $x$ axis with $v_x = +4.0\ \text{m/s}$ from $t = 0$ to $t = 3.0\ \text{s}$ , then $v_x = -2.0\ \text{m/s}$ from $t = 3.0\ \text{s}$ to $t = 8.0\ \text{s}$ . Find the displacement, the distance travelled, the average velocity and the average speed over the full 8.0 s. It starts at $x_0 = 0$ .

Convention: positive $x$ is already fixed by the graph, so a positive $v_x$ moves the object in the $+x$ direction and a negative $v_x$ moves it back.
First segment area. A rectangle of height $+4.0\ \text{m/s}$ and width $3.0\ \text{s}$ gives $+12\ \text{m}$ . It sits above the axis, so its contribution to the displacement is positive.
Second segment area. A rectangle of height $-2.0\ \text{m/s}$ and width $5.0\ \text{s}$ gives $-10\ \text{m}$ . It sits below the axis, so it subtracts.
Displacement, by EK 1.3.A.4.iii: add the signed areas. $\Delta x = +12 + (-10) = +2.0\ \text{m}$ . The object finishes 2.0 m to the right of where it started, at $x = +2.0\ \text{m}$ .
Distance, same areas without the signs: $\lvert +12 \rvert + \lvert -10 \rvert = 22\ \text{m}$ . The object went 12 m out and 10 m back.
Average velocity: $\vec{v}_{\text{avg}} = \frac{+2.0\ \text{m}}{8.0\ \text{s}} = +0.25\ \text{m/s}$ , in the $+x$ direction.
Average speed: $\frac{22\ \text{m}}{8.0\ \text{s}} = 2.75\ \text{m/s}$ .
Check the two pairs against each other. $22 \geq 2.0$ for the lengths and $2.75 \geq 0.25$ for the rates, and the ratio is 11 in both cases, because both pairs were divided by the same 8.0 s. The graph is one drawing and it contains both quantities; which one you get out depends entirely on whether you keep the signs.
Positions along the way, as a cross-check. The object reaches $x = +12\ \text{m}$ at $t = 3.0\ \text{s}$ , then moves left at $2.0\ \text{m/s}$ for 5.0 s to arrive at $12 - 10 = +2.0\ \text{m}$ , which matches the displacement found from the areas.

Displacement +2.0 m, distance 22 m, average velocity +0.25 m/s, average speed 2.75 m/s. The only difference between the two area calculations is whether the area below the axis is subtracted or counted positive, and that single choice changes the answer by a factor of 11.

Why work uses the displacement and friction uses the path

A 25 kg crate is pushed 4.0 m east across a level floor at constant speed, then pushed 4.0 m back west to its starting point, also at constant speed. The coefficient of kinetic friction is 0.20. Using $g = 9.8\ \text{m/s}^2$ , find the friction force, the work done by the push, the energy dissipated by friction, and the change in the crate's gravitational potential energy.

Convention: east is positive $x$ , up is positive $y$ . The crate stays on the floor, so $a_y = 0$ throughout.
Normal force, from the vertical second law with no vertical acceleration: $F_N = mg = (25\ \text{kg})(9.8\ \text{m/s}^2) = 245\ \text{N}$ .
Friction force. Sliding at constant speed, so $F_f = \mu F_N = (0.20)(245\ \text{N}) = 49\ \text{N}$ , opposing the motion on each leg and therefore reversing direction when the crate does.
Push force. Constant speed means zero acceleration, so the horizontal forces balance and the push is $49\ \text{N}$ , directed east on the way out and west on the way back.
Work done by the push, leg by leg. Each leg has the push parallel to the displacement of its point of application, so $\theta = 0$ and $W = Fd\cos 0 = (49\ \text{N})(4.0\ \text{m}) = 196\ \text{J}$ per leg. Total $W_{\text{push}} = 196 + 196 = 392\ \text{J}$ .
Now the error this example exists to expose. The crate's net displacement over the round trip is zero. Substituting that single net displacement into $W = Fd\cos\theta$ gives $W_{\text{push}} = 0$ , which would say the crate could be pushed there and back for free. The correct 392 J came from applying the relation once per leg, with each leg's own displacement and angle.
Energy dissipated by friction, from EK 3.2.A.4.iii, which uses the length of the path: $\Delta E_{\text{mech}} = F_f d = (49\ \text{N})(8.0\ \text{m}) = 392\ \text{J}$ . The full 8.0 m counts, with nothing cancelling, because friction reversed when the crate did and so kept opposing the motion.
Change in gravitational potential energy: $\Delta U_g = mg\Delta y = (25)(9.8)(0) = 0$ , because the crate ended at the height it started. Gravity is a conservative force and EK 3.2.A.1.ii makes this zero for any path that returns to its initial configuration.
Energy audit. The crate starts and finishes at rest, so $\Delta K = 0$ , and indeed $392\ \text{J}$ went in from the push and $392\ \text{J}$ came out as thermal energy. The books balance, which they could not if the push had done zero work.

Friction force 49 N, work done by the push 392 J, energy dissipated by friction 392 J, change in gravitational potential energy 0. The one quantity that used the net displacement, the potential-energy change, came out zero; the two that used the 8.0 m of path did not. Using the round trip's zero displacement in the work equation would have claimed the whole job cost nothing.

Frequently asked questions

What is the difference between distance and displacement?

Distance is the length of the path an object actually travelled, and it is a scalar with no direction. Displacement is the change in the object's position, drawn straight from the starting point to the ending point, and it is a vector with both a magnitude and a direction. The AP Physics 1 CED defines displacement at essential knowledge 1.2.A.2 as the change in an object's position, with the equation delta x equals x minus x nought, and names distance as an example of a scalar quantity at essential knowledge 1.1.A.3. The practical difference is that distance depends on the route and displacement depends only on the endpoints, so distance can never be smaller than the magnitude of the displacement.

Can displacement be negative?

Yes, in one dimension, where the sign is how the direction is expressed. The AP Physics 1 CED states at essential knowledge 1.1.A.3.ii that vector notation is not required for components along an axis and that in one dimension the sign of the component completely describes its direction, and at essential knowledge 1.1.B.1 that opposite directions are denoted by opposite signs. So a displacement of minus 5 metres means 5 metres in whichever direction you declared negative, and declaring that direction before you write any numbers is part of the answer. Distance can never be negative, because it is a path length.

Can displacement be zero when the distance travelled is not?

Yes, and this is the clearest demonstration that the two are different quantities. Any journey that finishes where it began has a displacement of exactly zero, because the change in position is zero, while the distance travelled is the full length of the route. A runner completing one lap of a 400 metre track has covered 400 metres of distance and has a displacement of zero. The same happens for a ball thrown straight up and caught at the launch height, and for one complete revolution of any circular motion. The reverse is impossible: the distance can never be smaller than the magnitude of the displacement.

How do you find displacement from a velocity time graph?

Take the area between the curve and the horizontal axis over the interval you want, counting area above the axis as positive and area below as negative, then add the signed pieces. The AP Physics 1 CED states this at essential knowledge 1.3.A.4.iii, describing the displacement as the area bounded by the function and the horizontal axis for the appropriate interval. To get the distance travelled from the same graph instead, take the absolute value of each area before adding, so nothing cancels. That single difference in procedure is the whole distinction between the two quantities, applied to a graph.

Is displacement the same as position?

No. A position is a single location, measured from an origin you chose. A displacement is the difference between two positions, so it needs a start and a finish. On the AP Physics 1 equation sheet the symbol key lists x as position and y as vertical position, and the displacement appears as the difference x minus x nought inside the third kinematic equation rather than as a symbol of its own. One consequence is that moving the origin changes every position in a problem but changes no displacement, because the same shift is subtracted out.

Does the d in the work equation mean distance or displacement?

It depends which relation you are using, which is why this catches people. The AP Physics 1 equation sheet prints work as W equals F parallel times d, and its symbol key reads d equals distance, but the CED prose at essential knowledge 3.2.A.3 describes the same quantity as the displacement of the point at which the force is exerted. Meanwhile essential knowledge 3.2.A.4.iii uses the same letter for the length of the path when equating the energy dissipated by friction to the friction force times that length. In practice: apply the work relation leg by leg using each leg's displacement, and use the total path length for friction dissipation.

Why does a car odometer measure distance rather than displacement?

Because it counts wheel rotations, and a wheel turns the same way whichever direction the car is pointing. Every metre of road adds to the total and nothing is ever subtracted, which is the defining behaviour of a distance. A displacement would require the odometer to know where the car started and where it is now, and to report a direction as well as a number, which is what a satellite navigation system does instead. Drive a hundred kilometres out and a hundred kilometres back and the odometer reads two hundred kilometres while the displacement is zero.