Calculus for Kinematics (Cambridge (CIE) O Level Additional Maths): Revision Note

The displacement, $x$ m, of a particle at $t$ seconds, is modelled by the function $x\left(t\right)=2t^3-27t^2+84t$.

Find expressions for $x\text{'}\left(t\right)$ and $x\text{'}\text{'}\left(t\right)$, and state what these expressions represent.

$x=2t^3-27t^2+84t$

$x\text{'}\left(t\right)$ means the same as $\frac{dx}{dt}$

$\begin{array}{l}x\text{'}\left(t\right)=6t^2-54t+84\\ x\text{'}\left(t\right)=6\left(t^2-9t+14\right)\end{array}$

$\mathit{x}\mathbf{\text{'}}\mathbf{\left(}\mathit{t}\mathbf{\right)}\mathbf{=}\mathbf{6}\mathbf{(}\mathit{t}\mathbf{-}\mathbf{2}\mathbf{)}\mathbf{(}\mathit{t}\mathbf{-}\mathbf{7}\mathbf{)}$
This expression represents the velocity of the particle.
Answers may or may not need to be factorised, depending on the question

$x\text{'}\text{'}\left(t\right)$ means the same as $\frac{d^{2}x}{d{t}^2}$

$x\text{'}\text{'}\left(t\right)=12t-54$

$\mathit{x}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{\left(}\mathit{t}\mathbf{\right)}\mathbf{=}\mathbf{6}\mathbf{(}\mathbf{2}\mathit{t}\mathbf{-}\mathbf{9}\mathbf{)}$
This expression represents the acceleration of the particle.

• Sketching the velocity-time graph can help you visualise the distances travelled using areas between the graph and the horizontal axis

A particle moving in a straight horizontal line has velocity $v$ ms^-1 at time $t$ seconds modelled by $v\left(t\right)=8t^3-12t^2-2t$.

a) Given that the initial position of the particle is at the origin, find an expression for its displacement from the origin at time $t$ seconds.

The integral of the velocity gives the displacement

$\begin{array}{l}s\left(t\right)=\int v\left(t\right)dt=\int \left(8t^3-12t^2-2t\right)dt\\ s\left(t\right)=2t^4-4t^3-t^2+c\end{array}$

"Initial" means when $t=0$ and "at the origin" means $s=0$

Substitute these values in to find $c$

$\begin{array}{l}0=2{\left(0\right)}^4-4{\left(0\right)}^3-{\left(0\right)}^2+c\\ c=0\end{array}$

$\mathit{s}\left(t\right)=2{\mathit{t}}^4-4{\mathit{t}}^3-{\mathit{t}}^2$

b) Find the displacement of the particle from the origin after the first five seconds of its motion.

To find the displacement after 5 seconds, integrate the velocity between 0 and 5

$\begin{array}{l}{\int }_0^{5}8t^3-12t^2-2tdt\\ {\left[2t^4-4t^3-t^2\right]}_0^5\\ =\left[2{\left(5\right)}^4-4{\left(5\right)}^3-{\left(5\right)}^2\right]-\left[2{\left(0\right)}^4-4{\left(0\right)}^3-{\left(0\right)}^2\right]\\ =725-0\end{array}$

$\mathbf{725}\mathbf{m}$

c) Explain why this value is not the same as the distance travelled during the first 5 seconds.

When finding the displacement, any negative displacements are taken into account. e.g. 3m - 1m = 2m. When finding the distance, we are looking for the total of all the areas, ignoring their direction/sign. e.g. 3m + 1m = 4m. This function for velocity has some regions underneath the x-axis, and some regions above the x-axis between 0 and 5 seconds, so the displacement and distance covered in this time will be different.