Speed, velocity and acceleration

A cyclist rides 3 km east, then 4 km west, and says "I covered 7 km." A GPS watching the same ride says "you ended up 1 km west of where you started." Both are true, and the gap between them is exactly the gap between distance and displacement — the seed from which speed, velocity and acceleration all grow. Speed answers "how much ground did I cover, and how fast?"; velocity answers "where did I end up, in what direction, and how fast?" Acceleration then asks how velocity itself is changing, moment to moment. Motion in a straight line looks simple until a direction reversal or a changing rate enters — exactly where careless answers lose marks, and where the exact definitions below make everything fall into place.

The core insight

Speed and its rate of increase (acceleration) are scalars built from distance and time, while velocity and acceleration proper are vectors built from displacement and time, and instantaneous values are always the limit of the average as the time interval shrinks to zero.

The core idea

Distance vs. displacement. Distance is the total path length travelled — always positive, always scalar, and it only grows as a body moves. Displacement is the straight-line change in position, from start point to end point — a vector, so it carries both magnitude and direction (or sign, on a single line). If a body reverses direction at any point, distance keeps climbing while displacement can shrink, stay flat, or even become zero.

Speed vs. velocity. Speed is how fast distance is being covered; velocity is how fast displacement is changing. The average versions over a time interval $\Delta t$ are

v_{\text{avg}} = \frac{\text{total distance}}{\text{total time}}, \qquad \vec v_{\text{avg}} = \frac{\Delta \vec x}{\Delta t} = \frac{\vec x_2 - \vec x_1}{t_2 - t_1},

where: total distance and total time are scalars, while $\Delta \vec x$ is the net displacement vector between the initial position $\vec x_1$ (at time $t_1$ ) and final position $\vec x_2$ (at time $t_2$ ). The two are equal in magnitude only for motion that never reverses direction; whenever the path doubles back, average speed is strictly greater than the magnitude of average velocity, since distance $\geq |\text{displacement}|$ always.

Instantaneous values. Shrinking the time interval to an instant gives the instantaneous velocity, the slope of the position–time graph at that point:

v = \lim_{\Delta t \to 0} \frac{\Delta x}{\Delta t} = \frac{dx}{dt}.

where: $\Delta x$ is the tiny displacement occurring in the tiny time $\Delta t$ , and the limit is the derivative of position with respect to time. Instantaneous speed is simply $|v|$ , the magnitude of the instantaneous velocity — so, unlike average speed and average velocity, the instantaneous pair are always linked by this absolute value, regardless of any earlier reversal. Motion is uniform if velocity stays constant (equal displacements in equal time intervals, zero acceleration) and non-uniform otherwise. Acceleration is the rate of change of velocity, again with average and instantaneous forms:

a_{\text{avg}} = \frac{\Delta v}{\Delta t}, \qquad a = \frac{dv}{dt} = \frac{d^2x}{dt^2}.

where: $\Delta v$ is the change in velocity over the time interval $\Delta t$ , and instantaneous acceleration is the derivative of velocity with time (equivalently, the second derivative of position). Acceleration is a vector with SI unit $\text{m/s}^2$ . Whether a body speeds up or slows down depends only on whether velocity and acceleration point the same way or opposite ways — not on whether acceleration itself is "positive" or "negative," which depends purely on the chosen sign convention.

Figure

A position time graph with a curved line whose slope steepens then flattens, a straight secant line marking average velocity between two points and a tangent line at one point marking instantaneous velocity

Derivation AInstantaneous velocity as the shrinking-interval limit of average velocity

The goal is to show, starting only from the everyday definition of average velocity over a finite time interval, exactly how narrowing that interval down to a single instant produces the calculus definition $v = dx/dt$ used throughout kinematics — and why this limiting process, not any new physical idea, is what turns a jerky "average over a stretch of time" into a smooth "velocity right now."

Step 1. Start with a body whose position is $x(t)$ , and pick two nearby instants: $t$ and $t + \Delta t$ , where $\Delta t$ is some small but non-zero time interval. The average velocity over this interval is, by definition, the net displacement divided by the elapsed time: $v_{\text{avg}} = \dfrac{x(t+\Delta t) - x(t)}{\Delta t} = \dfrac{\Delta x}{\Delta t}$ .

Step 2. This average velocity is only a single number for the whole interval — it hides whatever the body actually did in between (it could have sped up, slowed down, or briefly reversed). To pin down the velocity at the instant $t$ itself, the interval over which we average must be made shorter and shorter, so less variation is hidden inside it.

Step 3. Take the limit as $\Delta t \to 0$ . As the interval shrinks toward zero, both the numerator $\Delta x$ and the denominator $\Delta t$ shrink toward zero together, but their ratio settles down to a definite finite value — this is precisely the derivative of $x$ with respect to $t$ : $v = \displaystyle\lim_{\Delta t \to 0} \frac{x(t+\Delta t) - x(t)}{\Delta t} = \frac{dx}{dt}$ .

Step 4. Read the result physically. $v = dx/dt$ is not a different kind of velocity from $v_{\text{avg}}$ ; it is what $v_{\text{avg}}$ converges to as the averaging window collapses onto a single instant. Graphically, $v_{\text{avg}}$ over $[t, t+\Delta t]$ is the slope of the secant joining those two points on the position–time graph, and $v = dx/dt$ is the slope of the tangent at $t$ — the construction shown in the figure above. The same limiting argument, applied to $v(t)$ instead of $x(t)$ , gives instantaneous acceleration $a = dv/dt$ .

Worked example

The velocity of a particle moving along a straight line varies with time as $v(t) = 4t - t^2$ (SI units) for $0 \le t \le 5\ \text{s}$ . Find its acceleration at $t = 1\ \text{s}$ and $t = 3\ \text{s}$ , state whether it is speeding up or slowing down at each instant, and find the instant at which the particle is momentarily at rest.

Step 1. Differentiate to get acceleration as a function of time: $a(t) = \dfrac{dv}{dt} = 4 - 2t$ .

Step 2. Evaluate at $t = 1\ \text{s}$ : $v(1) = 4(1) - 1^2 = 3\ \text{m/s}$ (positive), and $a(1) = 4 - 2(1) = 2\ \text{m/s}^2$ (positive). Since $v$ and $a$ share the same sign, the particle is speeding up.

Step 3. Evaluate at $t = 3\ \text{s}$ : $v(3) = 4(3) - 3^2 = 3\ \text{m/s}$ (positive), and $a(3) = 4 - 2(3) = -2\ \text{m/s}^2$ (negative). Since $v$ is positive but $a$ is negative, the particle is slowing down, even though its speed is numerically the same $3\ \text{m/s}$ as at $t=1\ \text{s}$ .

Step 4. Momentary rest means $v(t) = 0$ : solve $4t - t^2 = 0$ , i.e., $t(4-t) = 0$ , giving $t = 0\ \text{s}$ or $t = 4\ \text{s}$ . Answer: $a(1\ \text{s}) = +2\ \text{m/s}^2$ (speeding up), $a(3\ \text{s}) = -2\ \text{m/s}^2$ (slowing down), and the particle is momentarily at rest at $t = 4\ \text{s}$ .

Competitive stretch

At what instant is the particle's speed maximum on $0 \le t \le 5\ \text{s}$ ? Maximum speed occurs either where $a = 0$ (a turning point of $v$ ) or at an endpoint. Setting $a(t) = 4-2t=0$ gives $t=2\ \text{s}$ , where $v(2)=4(2)-2^2=4\ \text{m/s}$ . Checking the endpoint $t=5\ \text{s}$ : $v(5)=4(5)-5^2=-5\ \text{m/s}$ , so $|v(5)|=5\ \text{m/s}$ — larger. The maximum speed is $5\ \text{m/s}$ , reached at the endpoint $t=5\ \text{s}$ , not at the interior turning point, a trap when a question asks for maximum speed rather than maximum velocity.

Common traps

Treating speed and velocity as interchangeable words: speed is a scalar that is never negative; velocity is a vector (or signed quantity on a line) that can be negative depending on the chosen positive direction.
Assuming average speed always equals the magnitude of average velocity — true only when motion never reverses direction along a straight line.
Believing negative acceleration always means "slowing down." It means slowing down only when velocity is positive; if velocity is also negative, a negative acceleration means the body is speeding up (in the negative direction).
Confusing "uniform motion" (constant velocity, zero acceleration) with "uniform acceleration" (constant acceleration, but velocity is still continuously changing).
Assuming maximum speed always occurs where acceleration is zero — as the competitive stretch shows, it can instead occur at the boundary of the time interval under consideration.

Ask the tutor: if the sign conventions for speeding up versus slowing down, or the secant-to-tangent limiting picture, still feel shaky, ask the AI tutor to walk through a few more velocity-time-function examples with you.

Why isn't average speed just the magnitude of average velocity, since both use time in the denominator?

Because the numerators are different kinds of quantities. Average speed divides total distance, the full path length, however winding. Average velocity divides net displacement, the straight-line distance from start to finish, which can be far shorter than the path (or even zero, if the body returns to its start). They only coincide when motion never reverses direction — any detour or U-turn makes distance strictly exceed the magnitude of displacement, so average speed becomes strictly larger.

If acceleration is negative, does that always mean the object is decelerating?

No — one of the most common exam slips. Whether a body speeds up or slows down depends on whether velocity and acceleration point the same way or opposite ways, not on the sign of acceleration alone. A negative acceleration means the body speeds up if its velocity is also negative, and slows down only if its velocity is positive. Always compare the signs of $v$ and $a$ together before concluding anything.

How is instantaneous velocity actually different from average velocity in practice, not just in the formula?

Average velocity smooths out everything that happened during the interval into one number — a car could have sped up, braked hard, and reversed briefly, and the average velocity over the whole trip would show none of that detail. Instantaneous velocity, by contrast, is what a speedometer-with-direction would read at one exact moment, obtained by shrinking the averaging interval to essentially zero. Graphically, average velocity is the slope of a secant on the position–time graph between two points, while instantaneous velocity is the slope of the tangent at one point — exactly what the derivative $dx/dt$ captures.

Speed, velocity and acceleration

The core idea

Worked example

Competitive stretch

Common traps

Before this

Where this leads