Kinematics using calculusEdexcel International A Level Further Maths: Revision notes

When the displacement $x$ of a particle on a line is a function of time, velocity is the rate of change of displacement and acceleration is the rate of change of velocity: $v=\frac{dx}{dt},\qquad a=\frac{dv}{dt}=\frac{d^2x}{dt^2}.$ Differentiating takes you from $x$ to $v$ to $a$; integrating takes you back. These formulae apply when the acceleration is not constant, so the suvat equations cannot be used. Example: $x=t^3-6t^2+9t$ gives $v=3t^2-12t+9$ and $a=6t-12$. The particle is at rest when $v=0$, i.e. $t=1$ and $t=3$.

• At rest: put $v=0$.
• Greatest or least velocity: put $a=0$ (then check ends of the interval).
• Acceleration zero gives the time of greatest or least velocity, not the time at rest.
• Speed is the magnitude $|v|$, so a negative velocity of $-9$ has speed $9$. Always evaluate at the end points of the time interval as well as at turning points when asked for the greatest value.

To go from acceleration to velocity, integrate and use an initial condition to find the constant: $v=\int a\,dt,\qquad x=\int v\,dt.$ Example: $a=6t-18$ with $v=24$ when $t=0$: $v=3t^2-18t+c$ and $c=24$. Integrating again with $x=0$ at $t=0$ gives $x=t^3-9t^2+24t$. Equations of the form $\frac{dx}{dt}=f(t)$ or $\frac{dv}{dt}=g(t)$ are solved directly by integration, as in P1 to P4.

The displacement is $x(t_2)-x(t_1)=\int_{t_1}^{t_2}v\,dt$ and can be positive, negative or zero. The total distance counts every movement as positive. If $v$ changes sign in the interval, split the motion at each time when $v=0$ and add the separate distances. Example: with $x=t^3-9t^2+24t$, $P$ turns at $t=2$ and $t=4$. For $0\leq t\leq5$: $x(2)=20$, $x(4)=16$, $x(5)=20$, so the distance is $20+4+4=28$ m, while the displacement is $20$ m.

If the position vector is $\mathbf{r}=x\mathbf{i}+y\mathbf{j}$, differentiate each component: $\mathbf{v}=\frac{d\mathbf{r}}{dt},\qquad\mathbf{a}=\frac{d\mathbf{v}}{dt}.$ Integrating $\mathbf{v}$ or $\mathbf{a}$ also works component by component, but each integration gives a vector constant, found from the initial conditions. The speed is $|\mathbf{v}|=\sqrt{v_x^2+v_y^2}$. Example: $\mathbf{r}=t^3\mathbf{i}+2t^2\mathbf{j}$ gives $\mathbf{v}=3t^2\mathbf{i}+4t\mathbf{j}$ and $\mathbf{a}=6t\mathbf{i}+4\mathbf{j}$. At $t=2$: $\mathbf{v}=12\mathbf{i}+8\mathbf{j}$, speed $\sqrt{208}=14.4$, and $\mathbf{a}=12\mathbf{i}+4\mathbf{j}$.