Calculus in kinematicsAQA A-Level Maths: Revision notes

The displacement $r$ of a particle moving in a straight line is its position measured from a fixed point $O$. It can be positive or negative, depending on which side of $O$ the particle is. Velocity is the rate of change of displacement and acceleration is the rate of change of velocity: $v=\frac{dr}{dt},\qquad a=\frac{dv}{dt}=\frac{d^2r}{dt^2}.$ Differentiate each power of $t$ in turn: $\frac{d}{dt}(t^n)=nt^{n-1}$. For $r=2t^3-9t^2+12t$ (metres, $t$ in seconds): $v=6t^2-18t+12$ and $a=12t-18$. When $t=3$: $r=9$ m, $v=12$ m s⁻¹, $a=18$ m s⁻². The units are m, m s⁻¹ and m s⁻².

Reversing differentiation gives displacement from velocity and velocity from acceleration: $v=\int a\,dt,\qquad r=\int v\,dt.$ Each indefinite integral has a constant of integration, found from the information given about the motion, such as the velocity or position at $t=0$. Worked example: $a=6t-4$, with $v=5$ when $t=0$ and $r=0$ when $t=0$. Then $v=3t^2-4t+c$ and $c=5$, so $v=3t^2-4t+5$. Integrating again, $r=t^3-2t^2+5t+d$ and $d=0$, so $r=t^3-2t^2+5t$. When $t=2$, $r=10$ m.

A particle is instantaneously at rest when $v=0$. Solve $v=0$ for $t$, then use the sign of $v$ on either side: if $v$ changes sign the particle reverses direction there, so $r$ has a turning value. Positive $v$ means moving in the positive direction, negative $v$ the opposite way. The velocity is greatest or least when $a=0$, as $a=\frac{dv}{dt}$. To show it is a maximum, show $\frac{da}{dt}<0$ there, or that $a$ changes from positive to negative. For $v=6t^2-18t+12=6(t-1)(t-2)$ the particle is at rest at $t=1$ and $t=2$.

Displacement is the change of position, with direction. Distance is the total length of path and is never negative. If the particle does not change direction, distance equals the size of the change in displacement. If it does, split the motion at each time when $v=0$ and add the sizes of each stage. Example: $v=3t^2-12t+9=3(t-1)(t-3)$ and $r=t^3-6t^2+9t$. In the first 3 s the particle moves from $r=0$ to $r(1)=4$, then back to $r(3)=0$. Displacement $=0$ m, but distance $=4+4=8$ m.

The equations of constant acceleration ($v=u+at$ and so on) hold only when $a$ is constant. If $a$ or $v$ is given as a function of $t$, use calculus. Check your results by substituting them back: differentiating your $r$ should reproduce the given $v$, and $r$ should have the right value at $t=0$. To compare two particles, write each displacement from the same origin and set them equal for "level". Reject any solution outside the stated time range, such as $t>6$ when the model is only valid for $0\le t\le6$.