Use calculus in kinematics: differentiate to move from displacement to velocity to acceleration, and integrate to reverse the process, for motion with variable acceleration.

What this dot point is asking

When acceleration is not constant, the suvat equations no longer apply, and CCEA GCSE Further Mathematics turns to calculus instead. You must differentiate displacement to get velocity and differentiate velocity to get acceleration, and reverse the chain by integrating, using the initial conditions to fix the constants. This is where the Pure unit's calculus meets the Mechanics unit, and it is a distinctive and well-rewarded topic.

Differentiating to go down the chain

The three quantities of motion form a chain linked by differentiation. Starting from a displacement that is a function of time, differentiating once gives velocity and differentiating again gives acceleration.

So if $s = 2t^3 - 5t$, then $v = 6t^2 - 5$ and $a = 12t$. Each derivative is found term by term using the power rule, exactly as in the Pure unit.

Special moments in the motion

Setting a derivative to zero pinpoints key instants. The particle is instantaneously at rest when $v = 0$, and the velocity is at a maximum or minimum when $a = \dfrac{dv}{dt} = 0$. These conditions answer common questions such as "when does the particle change direction?", which happens when the velocity passes through zero and changes sign.

For example, take $s = t^3 - 6t^2 + 9t$. Then $v = 3t^2 - 12t + 9 = 3(t - 1)(t - 3)$, which is zero at $t = 1$ and $t = 3$. Between these times the velocity is negative, so the particle moves backward, and outside them it moves forward; the two instants are where it changes direction. The maximum or minimum velocity is where $a = \dfrac{dv}{dt} = 6t - 12 = 0$, that is $t = 2$, the midpoint of the two rest instants. Identifying these special times turns a function of time into a full description of the motion, which is the kind of analysis CCEA expects.

Finding the distance travelled, as opposed to the displacement, also relies on these instants. Because the particle reverses direction, you must work out the displacement over each separate interval between rest instants and add their magnitudes, rather than simply substituting the start and end times. This distinction between distance and displacement is a frequent source of dropped marks.

Integrating to go up the chain

Reversing the chain uses integration, and each integration introduces a constant that an initial condition determines. Integrate acceleration to recover velocity, then integrate velocity to recover displacement.

Why this matters

Variable-acceleration kinematics shows calculus doing real physical work and is a clear demonstration of why differentiation and integration are inverse processes. It extends the constant-acceleration model to any motion described by a function of time, and the same chain underlies later study of dynamics. The skill of using initial conditions to find constants is exactly the technique met when recovering a function from its derivative in the Pure unit.