In short
Speed is total distance divided by total time — it tells you how fast, but not in which direction. Velocity is total displacement divided by total time — it tells you how fast and in which direction. Instantaneous velocity is the derivative v = \mathrm{d}x/\mathrm{d}t, and speed is the magnitude of velocity: \text{speed} = |v|. On a position–time graph, velocity is the slope.
An autorickshaw driver in Delhi picks you up at Connaught Place. The meter reads 8.4 km when you arrive at your destination in Karol Bagh. But if you could draw a straight line from Connaught Place to Karol Bagh on a map, it would be only about 4.2 km — the crow-flies distance. The autorickshaw wove through traffic, looped around one-way streets, and took a detour past Paharganj. It covered twice the straight-line distance.
So when someone asks "how fast was the ride?" — which distance do you use? The 8.4 km the odometer recorded, or the 4.2 km as the crow flies?
The answer is: both are useful, but they answer different questions. The first gives you speed. The second gives you velocity. The difference between them is the difference between "how far did you travel?" and "how far did you end up from where you started?" — the difference between distance and displacement, now measured per unit time.
Average speed — total distance over total time
Average speed is the simplest measure of motion. You take the total distance covered — every metre of the actual path, including every turn, every detour, every backtrack — and divide by the total time.
The autorickshaw covered 8.4 km in 28 minutes. Its average speed is:
Why divide by 28/60 instead of 28: because speed is in km/h, you must convert 28 minutes to hours. 28 min = 28/60 h ≈ 0.467 h.
Average speed is always positive (or zero, if nothing moved). It does not care about direction. An object that goes 5 km north and then 5 km south has covered 10 km of distance, and its average speed is 10 km divided by whatever time that took — it does not matter that the object ended up where it started.
Average speed is a scalar — a single number with no direction attached.
Average velocity — total displacement over total time
Average velocity asks a sharper question: not "how far did you travel?" but "how far did you end up from where you started, and in which direction?"
Here \Delta x = x_{\text{final}} - x_{\text{initial}} is the displacement — the straight-line distance from start to finish, with a sign (or direction) attached.
The autorickshaw moved 4.2 km roughly northwest from Connaught Place to Karol Bagh, and it took 28 minutes. Its average velocity is:
Why is this less than the average speed? Because displacement counts only the net effect of the motion — start to finish, in a straight line. The total path was twice as long as the displacement, so the average velocity is half the average speed.
Average velocity is a vector — it has both magnitude and direction. In one-dimensional motion (along a straight line), the "direction" reduces to a sign: positive means rightward (or forward), negative means leftward (or backward).
Here is the key difference, side by side:
| Average speed | Average velocity | |
|---|---|---|
| Uses | distance (total path length) | displacement (start-to-finish, straight line) |
| Type | scalar | vector |
| Can be negative? | never | yes (if displacement is negative) |
| Can be zero? | only if nothing moved | yes — if the object returns to the start |
The object that goes 5 km north and 5 km south has an average speed of 10 km / total time — but an average velocity of zero, because its displacement is zero. It ended up where it started. This is not a trick; it is a physically meaningful fact. The velocity tells you that despite all that running around, the net effect on position was nothing.
When speed and velocity differ — the circular track
Imagine a Delhi Metro train running on the Yellow Line's circular test track. It completes one full loop — say 2 km of track — in 4 minutes. What are its average speed and average velocity?
Average speed: the train covered 2 km of track, so 2 \text{ km} / (4/60) \text{ h} = 30 \text{ km/h}.
Average velocity: the train is back where it started. Displacement is zero. So average velocity = 0/4 = 0.
Why: this is the cleanest example of speed and velocity disagreeing. The train moved — it definitely was not sitting still — but its velocity over the full loop is zero because velocity tracks net displacement, not total distance.
For straight-line motion with no backtracking, distance equals displacement (in magnitude), so speed equals the magnitude of velocity. The two concepts diverge only when the path curves or the object reverses direction.
From average to instantaneous — velocity at a single moment
Average velocity tells you the overall rate of position change over a time interval. But physics often needs the velocity right now — at one specific instant. When a cricket fast bowler releases the ball, the speed gun does not measure the ball's average speed over the entire run-up. It measures the speed at the instant of release — the instantaneous speed.
How do you find velocity at a single instant? Start with the average velocity formula, and make the time interval smaller and smaller.
Pick a position function x(t) — say the position of a ball rolling along a track. At time t, the ball is at position x(t). At a slightly later time t + \Delta t, it is at x(t + \Delta t). The average velocity over this tiny interval is:
Why: this is just the average velocity formula applied to a small interval. The numerator is the displacement over the interval; the denominator is the duration.
Now imagine shrinking \Delta t toward zero — making the interval smaller and smaller. The average velocity over that shrinking interval approaches a single number: the instantaneous velocity at time t.
Why: this is the definition of a derivative. The instantaneous velocity is the derivative of the position function with respect to time. If you have studied differentiation, you recognise this as the slope of the x(t) curve at the point t.
The notation \mathrm{d}x/\mathrm{d}t means "the rate at which x is changing with respect to t, at this very instant." It is not a fraction (the \mathrm{d}'s are not numbers), but it behaves like one in many useful ways.
Instantaneous speed is the magnitude of instantaneous velocity:
If a ball is moving to the right at 5 m/s, its velocity is +5 m/s and its speed is 5 m/s. If it is moving to the left at 5 m/s, its velocity is -5 m/s but its speed is still 5 m/s. Speed does not care about direction; velocity does.
A numerical example — watching the limit form
Suppose a ball's position is given by x(t) = t^2 (in metres, with t in seconds). What is its instantaneous velocity at t = 3 s?
Compute the average velocity over smaller and smaller intervals around t = 3:
| Interval | \Delta t (s) | x(3 + \Delta t) (m) | \Delta x (m) | v_{\text{avg}} (m/s) |
|---|---|---|---|---|
| [3, 4] | 1 | 16 | 7 | 7.0 |
| [3, 3.5] | 0.5 | 12.25 | 3.25 | 6.5 |
| [3, 3.1] | 0.1 | 9.61 | 0.61 | 6.1 |
| [3, 3.01] | 0.01 | 9.0601 | 0.0601 | 6.01 |
| [3, 3.001] | 0.001 | 9.006001 | 0.006001 | 6.001 |
Why show this table: you can see the average velocity approaching 6 m/s as the interval shrinks. The limit is 6 m/s — that is the instantaneous velocity at t = 3 s.
You can verify this with calculus: v = \mathrm{d}x/\mathrm{d}t = 2t, so at t = 3, v = 6 m/s. The table shows the limit happening numerically; the derivative gives you the answer in one step.
Reading velocity from a position–time graph
A position–time graph has time t on the horizontal axis and position x on the vertical axis. Every point on the curve tells you where the object is at that moment. The slope of this curve at any point is the velocity at that instant.
This is not a metaphor — it is exactly the definition. Velocity is \mathrm{d}x/\mathrm{d}t, which is the slope of the x vs t graph. A steep upward slope means the object is moving fast in the positive direction. A gentle upward slope means it is moving slowly in the positive direction. A downward slope means it is moving in the negative direction. A horizontal section means the object has stopped — its velocity is zero.
Let us read the velocity from this graph at three moments:
- At t = 0: The curve rises steeply. The slope is positive and large. Velocity: v = 3 - t = 3 m/s.
- At t = 3 s: The curve reaches its peak — a horizontal tangent. The slope is zero. Velocity: v = 3 - 3 = 0. The object has momentarily stopped before reversing.
- At t = 5 s: The curve is falling. The slope is negative: v = 3 - 5 = -2 m/s. The object is moving in the negative direction.
Why does the object stop at t = 3? Because v(t) = 3 - t (the derivative of x = 3t - 0.5t^2), and v = 0 when t = 3. The position is at a maximum — the object has gone as far as it will go in the positive direction, and is about to turn around.
Slope as velocity — an interactive exploration
The figure below lets you drag a point along the position–time curve and watch the tangent line (whose slope is the velocity) rotate in real time. The readout shows both the position and the velocity at each instant.
Notice what happens when you drag past t = 3: the tangent line flips from upward-sloping to downward-sloping. The velocity passes through zero and becomes negative. The object has reversed direction. The graph does not "turn around" horizontally — the time axis always moves forward — but the position starts decreasing, which means the object is heading back toward where it came from.
The formal definitions
Let us collect everything in one place.
Average speed and average velocity
Average speed over a time interval \Delta t:
Average velocity over a time interval from t_1 to t_2:
Instantaneous velocity at time t:
Instantaneous speed at time t:
The SI unit of both speed and velocity is metres per second (m/s). You will also encounter km/h in everyday life: 1 \text{ m/s} = 3.6 \text{ km/h}.
Worked examples
Example 1: The Rajdhani Express — Delhi to Agra to Jaipur
A Rajdhani Express departs Delhi and travels 400 km to Agra in 2 hours. It then continues 200 km from Agra to Jaipur in 3 hours. The straight-line distance from Delhi to Jaipur is 260 km. Find: (a) the average speed for the entire trip, and (b) the magnitude of the average velocity.
Step 1. Find the total distance and total time.
Total distance = 400 + 200 = 600 km
Total time = 2 + 3 = 5 hours
Why: distance is additive along the path — the train covered 400 km on the first leg and 200 km on the second, so the total path length is 600 km. Time simply accumulates.
Step 2. Compute average speed.
Why: average speed uses total distance (the full rail path), not the straight-line displacement. This is 120 km/h, which is about 33.3 m/s.
Step 3. Compute the magnitude of average velocity.
The displacement is the straight-line distance from Delhi to Jaipur: 260 km.
Why: average velocity uses displacement — the net change in position, measured as a straight line from start to finish. The train covered 600 km of track, but it only ended up 260 km from where it started, so the average velocity is much less than the average speed.
Result: Average speed = 120 km/h. Magnitude of average velocity = 52 km/h.
What this shows: Average speed and average velocity can differ dramatically when the path is not a straight line. The more the path winds, the larger the gap between distance and displacement — and between speed and velocity.
Example 2: Finding when an object stops — instantaneous velocity from $x(t)$
A ball rolls along a track with position x(t) = t^3 - 6t (in metres, t in seconds). Find: (a) the instantaneous velocity at t = 2 s, and (b) the time(s) when the ball momentarily stops.
Step 1. Find the velocity function by differentiating x(t).
Why: the derivative of t^3 is 3t^2 (bring the exponent down, reduce it by 1), and the derivative of -6t is -6. The velocity function tells you the velocity at every instant, not just one.
Step 2. Evaluate v(t) at t = 2 s.
Why: substitute t = 2 into the velocity function. The result is positive, which means the ball is moving in the positive direction at t = 2 s. The speed at this instant is |v| = 6 m/s.
Step 3. Find when the ball stops. The ball stops when v = 0.
Why: v = 0 means the ball has zero velocity — it is momentarily at rest. Since t must be non-negative (time cannot run backward), the only solution is t = +\sqrt{2}. At this instant, the ball is at position x(\sqrt{2}) = (\sqrt{2})^3 - 6\sqrt{2} = 2\sqrt{2} - 6\sqrt{2} = -4\sqrt{2} \approx -5.66 m.
Step 4. Interpret the physics.
For 0 < t < \sqrt{2}: the velocity v = 3t^2 - 6 is negative (since 3t^2 < 6). The ball moves in the negative direction.
At t = \sqrt{2}: the velocity is zero. The ball stops momentarily.
For t > \sqrt{2}: the velocity is positive. The ball reverses and moves in the positive direction, accelerating as t increases.
Result: The instantaneous velocity at t = 2 s is 6 m/s (positive direction). The ball stops at t = \sqrt{2} \approx 1.41 s.
What this shows: Instantaneous velocity is the derivative of the position function. Setting v = 0 tells you exactly when an object reverses direction. The position–time graph confirms this — the minimum of the curve occurs at t = \sqrt{2}, precisely where the tangent is horizontal (slope = 0).
Common confusions
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"Speed and velocity are the same thing." They are the same only for motion in a straight line with no reversals. The moment the path curves, or the object doubles back, distance exceeds displacement and speed exceeds the magnitude of velocity. A runner on a circular track completes a lap with an average speed of 6 m/s but an average velocity of zero.
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"Average velocity = average of the velocities." This is a common algebraic trap. If a car drives at 40 km/h for 100 km and then at 60 km/h for 100 km, the average velocity is not (40 + 60)/2 = 50 km/h. The correct calculation uses total displacement over total time: 200/(100/40 + 100/60) = 200/(2.5 + 1.667) = 48 km/h. The average of the speeds gives 50 km/h, but the time-weighted average gives 48 km/h, because the car spent more time at the slower speed.
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"Velocity can never be zero if the object is moving." It can — but only for an instant. A ball thrown straight up has zero velocity at the very top of its flight. Its speed is zero at that instant. But it immediately starts falling back down — the velocity is zero at a single moment, not for a duration. The object does not "stop and wait" at the top; it passes through v = 0 the way a clock hand passes through 12 o'clock.
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"Negative velocity means the object is slowing down." Negative velocity means the object is moving in the negative direction — leftward, or downward, depending on your coordinate choice. It says nothing about whether the object is speeding up or slowing down. A car reversing and gaining speed has negative velocity that is increasing in magnitude. Whether the object is speeding up or slowing down is determined by acceleration, not by the sign of velocity.
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"The speedometer reads velocity." A car's speedometer reads instantaneous speed — it never shows a negative number, and it does not tell you which direction you are going. The GPS navigation, on the other hand, tracks your displacement over time and can compute velocity (with direction).
If you came here to understand the difference between speed and velocity, to use average and instantaneous versions, and to read velocity from a position–time graph, you have everything you need. What follows is for readers preparing for JEE who want the rigorous vector treatment and the calculus of non-uniform motion.
Vector velocity in two and three dimensions
Everything above was one-dimensional — motion along a line, with velocity being a signed number. In two or three dimensions, position is a vector \vec{r}(t) = x(t)\,\hat{i} + y(t)\,\hat{j} + z(t)\,\hat{k}, and velocity is the vector derivative:
Why: each component of position is differentiated independently. The velocity vector at any instant is tangent to the trajectory — it points in the direction the object is moving at that instant.
Speed in the vector case is the magnitude of the velocity vector:
For a projectile moving in 2D with x(t) = u\cos\theta \cdot t and y(t) = u\sin\theta \cdot t - \frac{1}{2}gt^2, the velocity components are v_x = u\cos\theta (constant) and v_y = u\sin\theta - gt (decreasing due to gravity). The speed at any instant is \sqrt{v_x^2 + v_y^2}, which is not constant even though the horizontal component is.
The relationship between distance and speed — the integral version
If velocity is the derivative of position, then position is the integral of velocity:
And the total distance covered is the integral of speed (the magnitude of velocity):
The distinction matters. If an object moves forward 10 m and then backward 3 m, the integral of velocity gives the displacement: +7 m. The integral of speed (the absolute value of velocity) gives the distance: 13 m.
Why average speed \geq magnitude of average velocity
This is always true, and the proof is a single line. By the triangle inequality for integrals:
The left side is |\text{displacement}|; the right side is the total distance. Dividing both sides by T:
Equality holds only when the object never reverses direction — when velocity does not change sign during the entire interval.
Speed as a function of arc length
In advanced mechanics (especially for constrained motion along curves), speed is sometimes written in terms of arc length s rather than time:
where s is the distance measured along the actual path. This is always non-negative, consistent with speed being the magnitude of velocity. The velocity vector can then be written as \vec{v} = v\,\hat{T}, where \hat{T} is the unit tangent vector to the path. This decomposition is the starting point for understanding centripetal and tangential acceleration in curved motion.
Where this leads next
- Acceleration — the rate at which velocity changes. Just as velocity is the derivative of position, acceleration is the derivative of velocity.
- Uniformly accelerated motion — when the acceleration is constant, you get the kinematic equations: v = u + at, s = ut + \frac{1}{2}at^2, and v^2 = u^2 + 2as.
- Graphs of motion — a detailed treatment of position–time, velocity–time, and acceleration–time graphs, and how to read each one.
- Position, distance, and displacement — the prerequisite article that defines the quantities that speed and velocity are built on.
- Calculus in physics: differentiation — the mathematical tool behind instantaneous velocity. If the derivative felt new, start here.