(12) A particle is moving in a straight line. Its position $x$ at time $t$ is given by \[ x=18-24 t+9 t^{2}-t^{3} \quad 0 \leqslant t \leqslant 5 \] (i) Find the velocity $v$ at time $t$ and the values of $t$ for which $v=0$. (ii) Find the position of the particle at those times. (iii) Find the total distance travelled in the interval $0 \leqslant t \leqslant 5$.

Step 1 ：First, we find the velocity function by taking the derivative of the position function with respect to time: \(v(t) = -3t^2 + 18t - 24\).

Step 2 ：Next, we find the values of \(t\) for which \(v=0\). We get \(t=2\) and \(t=4\).

Step 3 ：Then, we find the position of the particle at those times: \(x(2) = -2\) and \(x(4) = 2\).

Step 4 ：To find the total distance travelled, we find the position at \(t=0\) and \(t=5\), and then sum the absolute differences between consecutive positions: \(x(0) = 18\) and \(x(5) = -2\).

Step 5 ：Finally, the total distance travelled in the interval \(0 \leqslant t \leqslant 5\) is \(\boxed{28}\) units.