Calculus in kinematicsEdexcel A-Level Maths: Revision notes

For motion in a straight line, with displacement $s$ (or $r$) a function of time $t$: $v=\frac{ds}{dt},\qquad a=\frac{dv}{dt}=\frac{d^2s}{dt^2}.$ Velocity is the rate of change of displacement and acceleration is the rate of change of velocity. Differentiate term by term: $\frac{d}{dt}(t^n)=nt^{n-1}$. Example: $s=t^3-6t^2+9t$ gives $v=3t^2-12t+9$ and $a=6t-12$.

Integration reverses differentiation: $v=\int a\,dt,\qquad s=\int v\,dt.$ Use $\int t^n\,dt=\frac{t^{n+1}}{n+1}+c$. Every indefinite integral needs a constant of integration, found from initial conditions given in the question, such as 'at $t=0$, $v=2$' or 'at $t=0$ the particle is at $O$'. Example: $a=6t-4$ with $v=2$ at $t=0$. Then $v=3t^2-4t+c$, and $c=2$, so $v=3t^2-4t+2$. Integrating again with $s=0$ at $t=0$ gives $s=t^3-2t^2+2t$.

Instantaneously at rest means $v=0$: solve $\frac{ds}{dt}=0$. The particle may reverse direction at such a time. For a maximum or minimum velocity, set $a=0$. For a maximum or minimum displacement, set $v=0$ and substitute back into $s$. Example: $s=2t^3-9t^2+12t$ has $v=6t^2-18t+12=6(t-1)(t-2)$, so $v=0$ when $t=1$ or $2$, and $s=5$ or $4$ there.

Calculus gives signed quantities: a negative $v$ means motion in the negative direction. Displacement is $s$ at a time; distance travelled is the total path length. If $v$ changes sign, split the motion at the times when $v=0$ and add the magnitudes. If $v$ never changes sign, distance equals the magnitude of displacement from the start. To show a particle never changes direction, show that $v$ never equals zero, for example with a negative discriminant, and check one value of $v$ for the sign.