A car's speedometer shows 60 km/h. But how fast is it going right now — not on average, not over the last minute, but at this instant? That question forces you to build three distinct concepts: position, velocity, and acceleration. Here they are, derived not stated.

Position, Velocity, Acceleration — The Three Levels

A car travels 120 km in 2 hours. How fast was it going?

60 km/h, you say. And you are right — on average. But the car did not move at a perfectly constant 60 km/h for two hours. It accelerated from rest, cruised on the highway, slowed at a toll, sped up again. At some moments it was doing 90. At some moments, zero.

The number you calculated — 120 divided by 2 — tells you where the car ended up and how long it took. It says nothing about what the car was doing at any particular instant.

This is the gap that kinematics fills. Not that something moved, but how — at every instant, not just on average.

Position: the complete description

Before you can talk about motion, you need to answer a prior question: where is the object?

Position is the location of an object measured from a chosen reference point, in a chosen direction. The reference point is called the origin. The direction is called the positive direction. Both are chosen by you — the physics does not care which you pick, but once you pick, you must be consistent.

For motion along a line — a car on a highway, a ball thrown vertically — one coordinate is enough. Call it x. The position of the object at time t is written x(t): a function that gives one number for every moment in time.

$x(t)$

This function is the complete description of the motion. Everything else — velocity, acceleration — is extracted from it. If you know x(t) for all t, you know everything there is to know about where the object was, is, and will be.

Displacement — not the same as distance

The object starts at position x₁ at time t₁ and ends at position x₂ at time t₂.

The displacement is:

$\Delta x = x_2 - x_1$

It is a signed quantity. Positive if the object moved in the positive direction, negative if it moved in the opposite direction. The sign carries real information — it tells you which way.

Distance is the total length of the path actually travelled, with no regard for direction. Distance is always positive or zero.

If you walk 3 m east and then 3 m west, your displacement is zero. Your distance covered is 6 m. These are different numbers measuring different things. Confusing them is one of the most common errors in kinematics.

Displacement is a vector. Distance is a scalar. The moment you see a problem asking for "how far from the starting point," it is asking for the magnitude of displacement, not total distance — unless it explicitly says "total path length."

Velocity — the first level of change

The car moved from x₁ to x₂ in time Δt = t₂ − t₁. Its average velocity over that interval is:

$\bar{v} = \frac{\Delta x}{\Delta t} = \frac{x_2 - x_1}{t_2 - t_1}$

Average velocity is displacement divided by time — a signed quantity, like displacement. Average speed is total distance divided by time — always positive.

But average velocity does not tell you what was happening at any particular moment. To get that, you shrink the time interval.

Imagine the car at position x(t). After a tiny additional time Δt, it is at x(t + Δt). Its displacement in that interval is x(t + Δt) − x(t). Now let Δt become infinitesimally small — as you saw in M03, this is exactly what a derivative does:

$v(t) = \lim_{\Delta t \to 0} \frac{x(t + \Delta t) - x(t)}{\Delta t} = \frac{dx}{dt}$

This is instantaneous velocity: the derivative of position with respect to time. It is what the speedometer (with a direction attached) reads at a single moment. Not an average, not an approximation — the exact rate at which position is changing right now.

The magnitude of instantaneous velocity is instantaneous speed. This is what speedometers measure. Speed is always positive; velocity carries a sign.

Acceleration — the second level

Velocity itself can change. The car speeds up, slows down, or rounds a curve. The rate at which velocity changes is acceleration:

$a(t) = \frac{dv}{dt} = \frac{d^2x}{dt^2}$

Acceleration is the derivative of velocity, or equivalently, the second derivative of position. It is also a signed, directed quantity.

Here is where most students get confused.

Positive acceleration does not mean speeding up. It means the velocity is becoming more positive. If the object is moving in the negative direction (negative velocity) and the acceleration is positive, the object is actually slowing down.

Negative acceleration does not mean slowing down. "Deceleration" is an informal word, not a physics term. An object moving in the positive direction with negative acceleration is slowing down. An object moving in the negative direction with negative acceleration is speeding up (going faster in the negative direction).

The rule is this: if velocity and acceleration have the same sign, the object is speeding up. If they have opposite signs, the object is slowing down. The signs alone tell you everything — no separate concept of "deceleration" is needed.

The three levels together

This is the structure to hold:

Quantity	Symbol	How you get it	What it tells you
Position	x(t)	given directly	where the object is
Velocity	v(t) = dx/dt	differentiate position	how position is changing
Acceleration	a(t) = dv/dt	differentiate velocity	how velocity is changing

Each level is one derivative deeper than the one before. Position is the root. Velocity is its rate of change. Acceleration is the rate of change of that rate of change.

You can also read the table upward: if you know acceleration and want velocity, you integrate. If you know velocity and want position, you integrate again. This is the direction K02 will take — but only once you feel why these three levels exist and how they connect.

The key insight is that these are not three separate definitions to memorise. They are one idea unfolded three times. Nature gives you position. You ask "how fast is this changing?" — you get velocity. You ask again — you get acceleration. Beyond acceleration, you could ask again (the rate of change of acceleration is called jerk, and it matters in engineering) but for most of classical mechanics, three levels is enough.

Sign is not optional

Because velocity and acceleration are vectors, sign is part of the quantity — not a side note.

When you set up a problem, declare your positive direction explicitly. Usually: upward is positive for vertical problems, rightward is positive for horizontal ones. Then stay consistent.

A ball thrown upward has positive initial velocity. Gravity acts downward — it gives the ball a constant negative acceleration. The velocity decreases from positive, passes through zero at the top, becomes negative on the way down. The acceleration is constant and negative throughout, even at the top. Especially at the top: the ball momentarily has zero velocity, but its acceleration is not zero. It is g downward, as always.

This is a common error: confusing zero velocity with zero acceleration. They are independent. A ball at the top of its arc has v = 0 but a = −g. A car moving at constant velocity on a highway has v ≠ 0 but a = 0.

Note: When a position function x(t) is given:

> > 1. **Find velocity**: differentiate x(t) to get v(t) = dx/dt. > 2. **Find acceleration**: differentiate v(t) to get a(t) = dv/dt. > 3. **Find when object is stationary**: set v(t) = 0 and solve for t. > 4. **Find when object reverses direction**: also v(t) = 0, then check the sign of v just before and after. > 5. **Find displacement** over [t₁, t₂]: compute x(t₂) − x(t₁). Note: this can be zero even if the object moved. > 6. **Find total distance** over [t₁, t₂]: split the interval at every t where v = 0, add the absolute values of displacement on each sub-interval. > 7. **State direction**: always say "positive x direction" or "negative x direction" — never just a number without direction.

The common mistakes

Warning: Treating displacement and distance as the same. They are not. Displacement is signed and depends only on start and end positions. Distance depends on the entire path. If the object reversed direction, they differ.

> > **"Deceleration" as a separate concept.** There is no deceleration in physics — only acceleration. Whether the object is speeding up or slowing down depends on whether velocity and acceleration have the same sign, not on the sign of acceleration alone. > > **Zero velocity implies zero acceleration.** False. At the top of a vertical throw, v = 0 and a = −g simultaneously. Always treat them independently. > > **Average velocity equals instantaneous velocity.** Only true for uniform motion (constant velocity). In general, these are different numbers measuring different things. > > **Forgetting to check for direction reversal when computing distance.** If you compute distance as |x(t₂) − x(t₁)| without checking whether the object reversed, you will get the wrong answer whenever it did.

A worked example

A particle moves along the x-axis. Its position is given by:

$x(t) = t^3 - 6t^2 + 9t + 2 \quad \text{(in metres, t in seconds, } t \geq 0\text{)}$

Find: (a) the velocity at t = 2 s, (b) when the particle is momentarily at rest, (c) the acceleration at those moments, (d) the total distance covered in the first 4 seconds.

Step 1 — Find velocity.

$v(t) = \frac{dx}{dt} = 3t^2 - 12t + 9$

At t = 2: v(2) = 3(4) − 12(2) + 9 = 12 − 24 + 9 = −3 m/s

The particle is moving in the negative x direction at t = 2 s.

Step 2 — Find when v = 0.

$3t^2 - 12t + 9 = 0 \implies t^2 - 4t + 3 = 0 \implies (t-1)(t-3) = 0$

The particle is at rest at t = 1 s and t = 3 s.

Step 3 — Find acceleration at those moments.

$a(t) = \frac{dv}{dt} = 6t - 12$

At t = 1: a = 6(1) − 12 = −6 m/s² At t = 3: a = 6(3) − 12 = +6 m/s²

At t = 1, velocity is zero and acceleration is negative — the particle is about to move in the negative direction. At t = 3, velocity is zero and acceleration is positive — the particle is about to move in the positive direction. Both are turning points.

Step 4 — Total distance in [0, 4].

The particle reverses at t = 1 and t = 3. Split the interval.

Positions: x(0) = 2, x(1) = 1 − 6 + 9 + 2 = 6, x(3) = 27 − 54 + 27 + 2 = 2, x(4) = 64 − 96 + 36 + 2 = 6

Distances on each sub-interval:

[0, 1]: |6 − 2| = 4 m (moving positive)
[1, 3]: |2 − 6| = 4 m (moving negative)
[3, 4]: |6 − 2| = 4 m (moving positive)

Total distance = 4 + 4 + 4 = 12 m

Note: Displacement from t = 0 to t = 4 is x(4) − x(0) = 6 − 2 = 4 m. Total distance is 12 m. The two numbers differ because the particle reversed direction twice.

Twist variant

Same particle. A student computes the displacement from t = 0 to t = 4 as 4 m, then writes "the particle travelled 4 m." What is wrong, and what would the correct statement be?

The student found displacement — the net change in position. The particle actually covered 12 m of ground. The statement "travelled 4 m" uses the word "travelled" to mean distance, but plugged in displacement. Correct statement: "The particle has a displacement of 4 m in the positive x direction and covered a total distance of 12 m."

This conflation appears in roughly half of all first attempts at this type of problem.

What to remember

Position x(t) is the complete description of motion. Velocity is its derivative. Acceleration is velocity's derivative.

Displacement is signed and cares only about start and end. Distance is unsigned and cares about every step of the path.

Acceleration tells you how velocity is changing — not whether the object is speeding up. For that, compare the signs of v and a.

Zero velocity does not imply zero acceleration. They are independent quantities.

The next article, K02, takes a special case — constant acceleration — and extracts the four kinematic equations from first principles. Constant acceleration is rare in nature but ubiquitous in problems, which is why it deserves its own treatment. Once you have K01's framework, K02 is a single integration done twice.