The velocity–time graph gave us two treasures: the gradient was the acceleration, and the area underneath was the distance. Reading those off a straight line was easy — gradients of straight lines, areas of triangles and rectangles. But real motion is rarely so tidy. A sprinter, a bouncing ball, a car easing off the throttle: their graphs curve, and the acceleration is changing from one instant to the next.

The suvat equations you may have met only work when the acceleration is constant. Calculus lifts that restriction completely. It gives us the exact same two operations — gradient and area — but sharpened so they work on any curve, at any instant. Those two operations have names: differentiation (the instantaneous gradient) and integration (the exact area). Master them and you can describe motion of every kind, not just the straight-line special case.

Velocity is the gradient of the displacement–time graph

Displacement s tells you where the object is; velocity v tells you how fast that position is changing. On a displacement–time graph, "how fast the position is changing" is precisely the steepness of the curve — its gradient. Taken over a whole straight segment that is just \Delta s / \Delta t; taken at a single instant it is the gradient of the tangent line, and we write it with Leibniz's derivative symbol:

Read \dfrac{\mathrm{d}s}{\mathrm{d}t} as "the rate of change of s with respect to t" — the limit of \Delta s / \Delta t as the time interval shrinks to zero. The sign carries the direction: a positive gradient means s is growing (moving the positive way), a negative gradient means it is shrinking (moving back the other way), and a flat spot (zero gradient) is the instant the object is momentarily at rest.

Acceleration is the gradient of the velocity–time graph

Apply the very same idea one level up. Acceleration a is how fast the velocity is changing, so it is the gradient of the velocity–time graph:

And because v was itself the derivative of s, acceleration is the derivative of a derivative — the second derivative of displacement:

So a single displacement function s(t) holds the whole story: differentiate once for the velocity, differentiate again for the acceleration. Each step trades a graph for the graph of its slope.

Worked example: differentiate s(t) down to a(t)

Suppose an object moves so that its displacement in metres after t seconds is

Step 1 — velocity. Differentiate term by term with the power rule (bring the power down, drop it by one):

Step 2 — acceleration. Differentiate once more:

The acceleration is a plain 6\ \text{m/s}^2 — constant, because s was quadratic. At t = 0 the object is already moving: v = 6(0) + 2 = 2\ \text{m/s}, its starting velocity. By t = 4\ \text{s} it has sped up to v = 6(4) + 2 = 26\ \text{m/s}. Notice how the units drop out of each step: metres for s, metres per second for v, metres per second per second for a.

Going the other way: integration recovers the graph

Differentiation walks down the chain, s \to v \to a, taking gradients. Integration walks back up it, a \to v \to s, taking areas. The definite integral is exactly "the area under a curve", so:

The area under the acceleration–time graph is the change in velocity; the area under the velocity–time graph is the change in displacement — the very rule the velocity–time page used for triangles and rectangles, now good for any shape.

Worked example. A trolley starts from rest and its acceleration grows as a = 6t (in \text{m/s}^2). Integrate to get the velocity:

It starts from rest, so v = 0 when t = 0, which forces C = 0 and gives v = 3t^2. Integrate again for the displacement (starting at s = 0):

Differentiating s = t^3 straight back gives v = 3t^2 and then a = 6t — a neat check that the two operations really do undo one another. The constant C is not a nuisance: it is where the initial conditions (starting position, starting velocity) enter the problem.

suvat is just the constant-acceleration special case

Where do the familiar suvat equations come from? They are simply what integration gives when the acceleration a is a constant. Start there and integrate once, with initial velocity u:

That is the first suvat equation. Integrate the velocity to get the displacement (with s = 0 at t = 0):

The second suvat equation, falling straight out of the calculus. The lesson: suvat is not a separate rulebook to memorise — it is the special output of integration when, and only when, a is constant. The moment the acceleration varies, suvat breaks, but v = \int a\,\mathrm{d}t and s = \int v\,\mathrm{d}t carry on working.

When there is no formula, read the graph

Often you are handed a graph — from a data logger, a motion sensor, a crash test — with no equation at all. Calculus still tells you exactly what to do, because the two operations have direct graphical meanings:

• Need velocity from a displacement–time graph? Take the gradient — draw a tangent at the instant you care about and measure its slope.
• Need acceleration from a velocity–time graph? Again the gradient of the tangent.
• Need displacement from a velocity–time graph? Take the area — count squares under the curve, or split it into strips (that is exactly what a definite integral does in the limit).

The direction you travel decides which tool you reach for: gradient to go s \to v \to a, area to go a \to v \to s.

Watch the gradient become the velocity

Below is a displacement–time graph, s(t) = t^2, drawn in one colour, and its velocity graph v(t) = \frac{\mathrm{d}s}{\mathrm{d}t} = 2t drawn dashed. Drag the time slider: a tangent line rides along the curved displacement graph, and its slope is read off as the velocity. Watch how that slope always matches the height of the dashed velocity line directly below or above it — because the gradient of s at time t is the value of v at that time. Early on the curve is gentle (small velocity); later it steepens (larger velocity), exactly as v = 2t predicts.