Speed vs Velocity: What Is the Difference?

Speed is a scalar: how fast, with no direction. Velocity is a vector: how fast and in which direction. Both are measured in metres per second. Average speed comes from the distance along the path, average velocity from the displacement, so the two differ on any path that reverses or curves.

AP Physics: Unit 1 (topics 1.1 Scalars and Vectors in One Dimension, 1.2 Displacement, Velocity, and Acceleration, 1.3 Representing Motion). This comparison spans the first three topics of AP Physics 1 Unit 1, Kinematics, which carries 10 to 15 percent of the multiple-choice section over an estimated 12 to 17 class periods. EK 1.1.A.1 defines scalars as quantities described by magnitude only and vectors as quantities described by both magnitude and direction; EK 1.1.A.3 names distance and speed as scalars and position, displacement, velocity and acceleration as vectors. EK 1.2.B.2 defines average velocity as displacement divided by the time interval in which that displacement occurs, with the relevant equation v_avg = delta x / delta t. EK 1.2.B.4 states that an object is accelerating if the magnitude and/or direction of its velocity are changing. EK 1.3.A.4.i defines instantaneous velocity as the rate of change of position, equal to the slope of a tangent to a position against time graph, and EK 1.3.A.4.iii makes displacement the area under a velocity against time graph. The CED does not define average speed anywhere in the course framework and the phrase total distance does not appear in the document; the only discussion of average against instantaneous speed is in the Laboratory Investigations notes on photogates. 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

Speed is a magnitude. Velocity is a magnitude with a direction attached. That is the whole difference, and everything else on this page is a consequence of it.

The AP Physics 1 CED puts the two on opposite sides of its first classification. EK 1.1.A.1 says scalars are quantities described by magnitude only, while vectors are quantities described by both magnitude and direction. EK 1.1.A.3 then sorts the kinematic quantities into those two boxes by name: distance and speed are examples of scalar quantities, while position, displacement, velocity, and acceleration are examples of vector quantities.

So speed sits with distance, and velocity sits with displacement. That pairing is the useful part, because it tells you what each average is built from. An average speed is built from the length of the path the object actually travelled. An average velocity is built from the displacement, the straight arrow from start to finish. EK 1.2.B.2 gives the second one directly: average velocity is the displacement of an object divided by the interval of time in which that displacement occurs.

\vec{v}_{\text{avg}} = \frac{\Delta \vec{x}}{\Delta t}

The trap is that both quantities carry the same unit, metres per second, and in the simplest case they carry the same number as well. Nothing in the units or the arithmetic warns you which one a question wants. What warns you is the shape of the path.

This page is about telling them apart. The speed and velocity glossary entries define each on its own, and distance vs displacement handles the pair one level down, since that is where the difference actually originates.

Speed vs velocity, side by side

	Speed	Velocity
What it tells you	How fast	How fast, and which way
Scalar or vector	Scalar, by EK 1.1.A.3	Vector, by EK 1.1.A.3
Symbol	$v$ , or $\lvert \vec{v} \rvert$	$\vec{v}$ , or $v_x$ for a component along an axis
SI unit	$\text{m/s}$	$\text{m/s}$ , the same unit
Built from	The distance travelled along the path	The displacement, $\Delta \vec{x}$
Average over an interval	Path length divided by elapsed time	$\vec{v}_{\text{avg}} = \Delta \vec{x} / \Delta t$ , from EK 1.2.B.2
Can be negative	No	Yes in one dimension, where the sign is the direction
Value for a complete round trip	Greater than zero, as long as the object moved	Exactly zero, because $\Delta \vec{x} = 0$
Changes when the object turns at a steady rate	No	Yes, the direction changed
Instantaneous value	The magnitude of the instantaneous velocity	Rate of change of position, the slope of the tangent to an $x$ against $t$ graph, from EK 1.3.A.4.i
Defined in the CED	Named as a scalar in EK 1.1.A.3	Averaged form defined in EK 1.2.B.2
On the AP Physics 1 equation sheet	The symbol key has one entry covering both: $v$ = velocity or speed	The same single entry

Two rows are worth reading together. The unit row and the round-trip row. Identical units are why the two words get used interchangeably in conversation, and the round trip is where that habit produces two different numbers from the same journey. Every worked example below is a version of that.

The case that separates them: a path that doubles back

Here is the smallest situation in which the two quantities disagree. A runner goes 300 m east in 100 s, then turns around and comes 100 m back west in 50 s. Take east as positive.

Quantity	Value	Where it came from
Distance travelled	$300 + 100 = 400\ \text{m}$	Add up the path, ignoring direction
Displacement	$+300 - 100 = +200\ \text{m}$	Signed sum, east positive
Elapsed time	$100 + 50 = 150\ \text{s}$	Add the intervals
Average speed	$400 / 150 = 2.67\ \text{m/s}$	Path length over time
Average velocity	$200 / 150 = 1.33\ \text{m/s}$ east	$\Delta \vec{x} / \Delta t$

Same journey, same clock, two different numbers, and the average speed is exactly twice the average velocity here. Nothing went wrong. The two quantities were asking different questions, and the 100 m of backtracking counts positively toward one and negatively toward the other.

Push the same idea to its limit and the runner comes all the way home. Then the displacement is zero, so the average velocity is exactly zero however hard the runner ran, while the average speed is whatever the lap distance over the lap time comes to. A 400 m lap in 80 s gives an average speed of $5.0\ \text{m/s}$ and an average velocity of $0$ . That is not a paradox and it is not a trick question: it is EK 1.2.B.2 applied to $\Delta \vec{x} = 0$ .

The rule the two numbers obey. The distance travelled is never smaller than the magnitude of the displacement, because the straight line between two points is the shortest path between them. So

\text{average speed} \geq \lvert \vec{v}_{\text{avg}} \rvert

always, with equality only when the object never reverses direction. If you compute an average speed smaller than the magnitude of your average velocity, you have made an arithmetic error, and this inequality is the fastest way to catch it.

Constant speed is not constant velocity

This is the version of the distinction that appears most often in a multiple-choice question, because it can be asked without any numbers at all.

EK 1.2.B.4 is the relevant line: an object is accelerating if the magnitude and/or direction of the object's velocity are changing. Read the "and/or". Changing the magnitude counts. Changing the direction counts. Changing only the direction still counts.

So a car going round a circular track at a steady 20 m/s has:

constant speed, since the magnitude never changes,
continuously changing velocity, since the direction changes at every instant,
and therefore a nonzero acceleration the whole way round, which is why circular motion belongs to dynamics at all.

A question that says "the car moves at constant speed, is it accelerating?" is testing exactly one thing, and the answer is yes. A question that says "the car moves at constant velocity" has told you something much stronger: straight line, unchanging rate, zero acceleration, and by EK 2.4.A.3 a net force of zero.

That is also the reason "constant velocity" is worth reading as three claims rather than one. It means the magnitude is fixed and the direction is fixed. Miss the second half and you will call a satellite in a circular orbit a constant-velocity object, which then makes the centripetal force impossible to justify. The procedure for that force lives in centripetal force; the point here is only that constant speed and constant velocity are different premises.

The definition the CED never gives you

This is worth knowing before an exam, because it explains a gap you might otherwise think you had missed.

Open almost any textbook and average speed arrives as a formula: total distance divided by total elapsed time. That formula does not appear anywhere in the AP Physics 1 CED. The phrase "total distance" does not occur in the document. The words "average speed" occur only in the Laboratory Investigations section, in the notes on what photogates measure, and nowhere in the course framework itself.

What the framework does instead is this:

EK 1.1.A.3 names speed as an example of a scalar quantity, alongside distance. That is a classification, not a formula.
EK 1.2.B.2 defines average velocity as displacement over the time interval, and EK 1.2.B.3 defines average acceleration as change in velocity over the time interval. Those are the two averages the framework defines.
EK 1.3.A.4.i defines instantaneous velocity as the rate of change of position, equal to the slope of a line tangent to a point on a graph of position against time.

So the CED builds the whole quantitative side of Unit 1 on velocity, treats speed as a magnitude you can name when you need it, and never turns average speed into an equation. That does not make average speed unexaminable, since a question can perfectly well hand you a path length and a time. It does mean that when a problem gives you a curved or reversing path, the quantity the framework has actually equipped you to compute is the average velocity, and the average speed is a separate arithmetic you do by hand from the path.

The one place the CED discusses the distinction directly is instructive about why it bothers. In the Laboratory Investigations notes on photogates, it says that photogates can be said to determine speed, but that depending on the configuration, the speed being measured may be considered to be equal to the object's instantaneous speed at that single location, or the average speed between two different photogates, and then adds: "In some applications, this distinction is crucial to obtaining the appropriate value." That is a statement about equipment, and it is also the cleanest sentence in the document about why any of this matters.

Where it costs a mark

Each of these is a specific scoring event.

Reporting an average velocity with no direction. Velocity is a vector. In one dimension a sign is enough, since EK 1.1.B.1 says opposite directions are denoted by opposite signs, but a bare positive number with no stated convention is incomplete. Declare your axis, then use it.
Reporting a negative speed. A speed cannot be negative, because it is a magnitude. If your arithmetic produces $-3.0\ \text{m/s}$ and the question asked for speed, the answer is $3.0\ \text{m/s}$ and the minus sign belongs to a velocity.
Dividing the path length by the time and calling it the average velocity. This is the error the worked examples are built around. It gives a number that is too large whenever the object reversed, and it can never be zero, so it cannot reproduce the round-trip answer.
Taking the magnitude of the average velocity to be the average speed. They agree only when the motion never reverses. Over the runner's out-and-back trip above, one is $2.67\ \text{m/s}$ and the other is $1.33\ \text{m/s}$ .
Averaging speeds instead of using total path over total time. Two legs at different speeds do not give the mean of the two speeds unless the legs took equal times. Worked example three shows a case where the naive mean is wrong.
Calling constant-speed circular motion unaccelerated. EK 1.2.B.4 counts a change of direction as acceleration.
Saying the object is at rest at the top of a vertical throw because the speed is zero there. The speed is zero for an instant; the acceleration is still $9.8\ \text{m/s}^2$ downward, so the velocity is changing through zero rather than sitting at it.
Reading a velocity off the wrong feature of a graph. On a position against time graph, the average velocity over an interval is the slope of the straight line joining the two endpoints, while the instantaneous velocity is the slope of the tangent, per EK 1.3.A.4.i. On a velocity against time graph, the displacement is the signed area under the curve, per EK 1.3.A.4.iii, so area below the axis subtracts. Add the areas without their signs and you have computed a distance, not a displacement, and dividing that by the time gives an average speed.

When they coincide, and why that lulls you

The two quantities give the same number whenever the object moves along a straight line and never reverses direction. Under that condition the path length and the magnitude of the displacement are the same measurement, so both averages divide the same number by the same time.

That covers almost everything in the first weeks of a course. A cart on a track released from rest. A ball dropped from a window. A car braking to a stop. In each of those the two words are interchangeable and nobody is punished for treating them that way.

Here is what breaks the coincidence, in roughly the order it appears on a syllabus:

Anything that comes back. A ball thrown straight up and caught, a cart bouncing off a spring, a pendulum. The moment the velocity changes sign, the path length outruns the displacement.
Anything two-dimensional. A projectile, a circle, a car turning a corner. The displacement is now the hypotenuse of the trip and the path is longer than it.
Anything where the question says "average". That word is the flag. It forces you to decide which of the two quantities is wanted, and the answer is in whether the question names a distance or a displacement.

The habit worth building is to write down the displacement and the path length as two separate labelled numbers before computing anything. On a straight one-way trip they will be equal and the extra line costs you nothing. On any other trip the two numbers are visibly different and you cannot then divide the wrong one by mistake.

What the equation sheet gives you, and what it does not

The AP Physics 1 equation sheet is a velocity sheet, and its treatment of speed is a single line in the symbol key.

In the symbol key, one entry covers both quantities: $v$ = velocity or speed. The sheet also lists $d$ = distance and $x$ = position separately, which is the same scalar-and-vector pairing as EK 1.1.A.3, applied to the letters. So the sheet distinguishes distance from position but uses one letter for velocity and speed, and the burden of knowing which one a given line means falls on you.

In the equations, the three kinematic relations are written for a velocity component along an axis:

v_x = v_{x0} + a_x t

x = x_0 + v_{x0} t + \frac{1}{2} a_x t^2

v_x^2 = v_{x0}^2 + 2 a_x (x - x_0)

Every $v$ in those is a signed component, not a speed, and every $x$ is a position. Feed a path length into the third one in place of $x - x_0$ and it will return a wrong answer without complaining. EK 1.3.A.2 lists exactly these three and notes that they are written for the $x$ direction but can be used in any single dimension as appropriate. The routine for using them is in kinematic equations.

What the sheet does not print is any definition of an average. There is no $\vec{v}_{\text{avg}} = \Delta \vec{x} / \Delta t$ line, no average-acceleration line, and no average-speed line. The CED lists the average-velocity relation as a relevant equation under EK 1.2.B.2, but it did not make the printed sheet, so on exam day you write it from the definition. Worth contrasting with power, where the sheet does print $P_{\text{avg}} = W/\Delta t = \Delta E/\Delta t$ : the sheet is not systematically averse to averages, it just does not carry this one.

The full sheet is here. For the CED framing of these topics, Topic 1.1 covers the scalar and vector split, Topic 1.2 covers the averages, and Topic 1.3 covers reading them off graphs. Average vs instantaneous velocity takes the other cut through the same material.

Out and back: one journey, two numbers

A runner travels 300 m east in 100 s, then turns and travels 100 m west in 50 s. Find the distance travelled, the displacement, the average speed and the average velocity over the whole 150 s. Then find all four quantities for the case where the runner instead completes a 400 m circular lap in 80 s. Take east as positive.

Declare the convention: east is the positive direction throughout, so the westward leg carries a minus sign and the eastward leg does not.
Distance for the out-and-back trip. Distance is a scalar and ignores direction, so add the path lengths: $300\ \text{m} + 100\ \text{m} = 400\ \text{m}$ .
Displacement. This is a signed sum, so $\Delta x = +300\ \text{m} + (-100\ \text{m}) = +200\ \text{m}$ , meaning 200 m east of the start.
Average speed. Path length over elapsed time: $400\ \text{m} / 150\ \text{s} = 2.667\ \text{m/s}$ , which is $2.67\ \text{m/s}$ to three significant figures.
Average velocity. Displacement over elapsed time, from EK 1.2.B.2: $\vec{v}_{\text{avg}} = +200\ \text{m} / 150\ \text{s} = +1.333\ \text{m/s}$ , or $1.33\ \text{m/s}$ east.
Check the ratio. $2.667 / 1.333 = 2.00$ , and the reason is visible in the numbers: the path is 400 m while the displacement is 200 m, and both were divided by the same 150 s. The time never causes the difference; the geometry does.
Now the lap. Distance travelled is the full $400\ \text{m}$ of track. Displacement is $0$ , because the runner finishes where they started.
So the lap gives average speed $= 400\ \text{m} / 80\ \text{s} = 5.0\ \text{m/s}$ and average velocity $= 0\ \text{m} / 80\ \text{s} = 0$ . Not approximately zero: exactly zero, for any lap, at any pace.
Verify the inequality. For the out-and-back trip $2.67 \geq 1.33$ ; for the lap $5.0 \geq 0$ . The average speed can never come out below the magnitude of the average velocity, because a path can never be shorter than the straight line across it.

Out and back: distance 400 m, displacement +200 m (east), average speed 2.67 m/s, average velocity 1.33 m/s east. The lap: distance 400 m, displacement 0, average speed 5.0 m/s, average velocity 0. The lap is the extreme case of the same effect. A runner can be exhausted and still have an average velocity of exactly zero, because average velocity measures where you ended up and not how hard you worked.

Constant speed, changing velocity, nonzero acceleration

A car drives around a circular track of radius 50 m at a constant speed of 20 m/s. Find its centripetal acceleration, then find its average speed and the magnitude of its average velocity over half a lap and over one full lap.

The speed is constant at $20\ \text{m/s}$ , so the average speed over any interval of this motion is also $20\ \text{m/s}$ . That is the easy half of the question.
The velocity is not constant, because its direction changes continuously. By EK 1.2.B.4 the car is accelerating, and the acceleration is centripetal: $a_c = \frac{v^2}{r} = \frac{(20\ \text{m/s})^2}{50\ \text{m}} = \frac{400}{50} = 8.0\ \text{m/s}^2$ , directed toward the centre.
Half a lap, geometry first. The path is half the circumference, $\pi r = \pi (50\ \text{m}) = 157.08\ \text{m}$ . The displacement is the diameter, $2r = 100\ \text{m}$ , pointing from the start straight across to the finish.
Half a lap, timing. At $20\ \text{m/s}$ the car covers $157.08\ \text{m}$ in $t = 157.08 / 20 = 7.854\ \text{s}$ .
Half a lap, the two averages. Average speed $= 157.08 / 7.854 = 20.0\ \text{m/s}$ , as it must be. Magnitude of average velocity $= 100 / 7.854 = 12.73\ \text{m/s}$ , so $12.7\ \text{m/s}$ to three significant figures.
Note what the ratio is. $12.73 / 20.0 = 0.637$ , which is $2/\pi$ . The factor is pure geometry: the diameter is $2/\pi$ of the semicircular arc, and both were divided by the same time.
One full lap. Path $= 2\pi r = 314.16\ \text{m}$ , time $= 314.16 / 20 = 15.71\ \text{s}$ , displacement $= 0$ . Average speed $= 20.0\ \text{m/s}$ ; average velocity $= 0$ .
Collect the point. Across all three intervals the speed never changed by a single digit, the acceleration was $8.0\ \text{m/s}^2$ throughout, and the average velocity ran from $20\ \text{m/s}$ over a tiny arc down to $0$ over a full lap. Speed and velocity are not the same quantity in different clothes.

Centripetal acceleration 8.0 m/s $^2$ toward the centre, the whole way round. Half a lap: average speed 20.0 m/s, magnitude of average velocity 12.7 m/s, a ratio of $2/\pi$ . Full lap: average speed 20.0 m/s, average velocity 0. Constant speed and zero net force are different claims, and this motion has the first without the second.

A ball thrown straight up, where the average speed is not the mean of two speeds

A ball is thrown straight up from a hand at $14\ \text{m/s}$ and caught at the same height on the way down. Using $g = 9.8\ \text{m/s}^2$ , find the maximum height, the total flight time, and the average speed and average velocity over the upward half and over the whole flight. Then check what the mean of the starting and ending speeds would have given.

Declare the convention: up is positive, so the acceleration is $a_y = -9.8\ \text{m/s}^2$ for the entire flight, including at the top.
Time to the top. At the top $v_y = 0$ , so $0 = 14 + (-9.8)t$ gives $t_{\text{up}} = 14 / 9.8 = 1.4286\ \text{s}$ .
Maximum height. Use $v_y^2 = v_{y0}^2 + 2a_y \Delta y$ with $v_y = 0$ : $0 = 196 + 2(-9.8)\Delta y$ , so $\Delta y = 196 / 19.6 = 10.0\ \text{m}$ exactly.
Total flight time. The motion is symmetric about the top, so $t_{\text{total}} = 2(1.4286) = 2.857\ \text{s}$ , and the ball returns with a speed of $14\ \text{m/s}$ directed downward.
Upward half. Path length $= 10.0\ \text{m}$ , displacement $= +10.0\ \text{m}$ , because nothing reversed. Average speed $= 10.0 / 1.4286 = 7.00\ \text{m/s}$ ; average velocity $= +10.0 / 1.4286 = 7.00\ \text{m/s}$ upward. On this leg the two agree, which is the coincidence condition in action.
Whole flight. Path length $= 10.0 + 10.0 = 20.0\ \text{m}$ , displacement $= 0$ , since the ball is caught at the launch height. Average speed $= 20.0 / 2.857 = 7.00\ \text{m/s}$ ; average velocity $= 0 / 2.857 = 0$ .
Now the naive shortcut. The starting speed is $14\ \text{m/s}$ and the ending speed is $14\ \text{m/s}$ , so averaging the two speeds gives $14\ \text{m/s}$ , which is double the correct $7.00\ \text{m/s}$ . Averaging endpoint values works for a signed velocity under constant acceleration, not for a speed that dips to zero in between.
One more thing at the top. There the speed is $0$ and the velocity is $0$ , but the acceleration is still $9.8\ \text{m/s}^2$ downward. Zero speed for one instant is not rest; it is the moment the velocity passes through zero on its way to negative.

Maximum height 10.0 m, flight time 2.857 s. Upward half: average speed 7.00 m/s and average velocity 7.00 m/s upward, equal because the ball never reversed. Whole flight: average speed 7.00 m/s and average velocity 0. The mean of the two endpoint speeds gives $14\ \text{m/s}$ , twice the true average speed, because a speed cannot be averaged from its endpoints when it passed through zero on the way.

Frequently asked questions

What is the difference between speed and velocity?

Speed says how fast something is moving. Velocity says how fast and in which direction. The AP Physics 1 CED puts them on opposite sides of its first classification: essential knowledge 1.1.A.1 defines a scalar as a quantity described by magnitude only and a vector as one described by both magnitude and direction, and essential knowledge 1.1.A.3 names speed as a scalar and velocity as a vector. Both are measured in metres per second, which is why they are so easily confused. The practical consequence is that an average speed is computed from the length of the path travelled, while an average velocity is computed from the displacement, so the two give different numbers for any path that changes direction.

Can speed be negative?

No. Speed is a magnitude, and magnitudes are never negative. A velocity in one dimension can be negative, because the sign carries the direction: essential knowledge 1.1.B.1 states that in a given one-dimensional coordinate system, opposite directions are denoted by opposite signs. So a ball falling at minus 12 metres per second in an upward-positive system has a speed of 12 metres per second. If a calculation gives you a negative number and the question asked for a speed, report the magnitude and keep the sign for the velocity.

Is average speed the same as the magnitude of average velocity?

Only when the object never reverses direction. Average speed uses the length of the path actually travelled, and average velocity uses the displacement, which is the straight-line change in position. A path is never shorter than the straight line across it, so the average speed is always greater than or equal to the magnitude of the average velocity. A runner who goes 300 metres east and then 100 metres back west in 150 seconds has an average speed of 2.67 metres per second and an average velocity of 1.33 metres per second east. For a complete round trip the average velocity is exactly zero while the average speed is not.

Can an object have constant speed and still be accelerating?

Yes, and this is the most commonly examined form of the distinction. Essential knowledge 1.2.B.4 in the AP Physics 1 CED states that an object is accelerating if the magnitude and or the direction of its velocity are changing. A car going round a circular track at a steady 20 metres per second has an unchanging speed but a continuously changing direction, so its velocity is changing and it is accelerating the whole way round. Its acceleration is centripetal, of magnitude v squared over r, and it points toward the centre of the circle. Constant velocity is the much stronger statement: fixed magnitude and fixed direction, therefore zero acceleration and zero net force.

Why is average velocity zero for a round trip when the average speed is not?

Because average velocity is defined from the displacement rather than from the distance travelled. Essential knowledge 1.2.B.2 says average velocity is the displacement of an object divided by the interval of time in which that displacement occurs. A round trip ends where it began, so the displacement is exactly zero, and zero divided by any time is zero. The average speed uses the path length instead, which is the full lap, so it stays positive. The two answers are not in conflict: they are answers to two different questions, one about where you ended up and one about how much ground you covered.

Is instantaneous speed the same as the magnitude of instantaneous velocity?

Yes. At a single instant there is no interval for a path to double back over, so the two agree exactly and always. The distinction between speed and velocity only produces different numbers when you average over an interval. This is why a speedometer reading and the magnitude of the car's instantaneous velocity are the same number, while a trip computer's average speed and the trip's average velocity can differ enormously. The AP Physics 1 CED defines the instantaneous quantity in velocity terms at essential knowledge 1.3.A.4.i, as the rate of change of position, equal to the slope of a line tangent to a point on a graph of position against time.

Does the AP Physics 1 equation sheet give a formula for average speed?

No, and it does not give one for average velocity either. The sheet prints the three constant-acceleration kinematic equations, all written for a signed velocity component, and its symbol key carries a single entry reading v equals velocity or speed. The average-velocity relation, displacement divided by time interval, appears in the CED as a relevant equation under essential knowledge 1.2.B.2 but is not on the printed sheet, so you write it from the definition. The formula usually taught for average speed, total distance divided by total time, does not appear in the CED at all: the phrase total distance is absent from the document, and the words average speed occur only in the section on laboratory equipment.