Answer: \(t=\frac13\); moving towards \(O\) at \(t=0\); total distance \(=\frac{280}{27}=10.37\ldots\text{ m}\).

Use differentiation to find the gradient information needed. Stationary points occur when the derivative is zero, and tangents or normals are found from the gradient at the given point.

The displacement is \(x=t^3+t^2-t+8\). The velocity is

\(v=\frac{\mathrm dx}{\mathrm dt}=3t^2+2t-1.\)

The particle changes direction when \(v=0\):

\(3t^2+2t-1=0\), so \((3t-1)(t+1)=0.\)

Since \(t\geqslant0\), the time is \(t=\frac13\).

At \(t=0\), \(x=8\), so the particle is to the positive side of \(O\). Also \(v=-1\), so it is moving in the negative direction. Hence it is moving towards \(O\).

At \(t=\frac13\),

\(x=\left(\frac13\right)^3+\left(\frac13\right)^2-\frac13+8=\frac{211}{27}.\)

At \(t=2\),

\(x=8+4-2+8=18.\)

From \(t=0\) to \(t=\frac13\), the distance travelled is

\(8-\frac{211}{27}=\frac{5}{27}.\)

From \(t=\frac13\) to \(t=2\), the distance travelled is

\(18-\frac{211}{27}=\frac{275}{27}.\)

The total distance is therefore

\(\frac{5}{27}+\frac{275}{27}=\frac{280}{27}=10.37\ldots\text{ m}.\)

This gives the required answer: \(t=\frac13\); moving towards \(O\) at \(t=0\); total distance \(=\frac{280}{27}=10.37\ldots\text{ m}\).