Answer: \(t=\frac12\), distance \(=3.88\text{ m}\), acceleration \(=-\frac{20}{(2t+1)^2}\), and at \(t=4.5\), acceleration \(=-0.2\text{ m s}^{-2}\).

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.

Differentiate displacement to find velocity:

\(v=\frac{\mathrm ds}{\mathrm dt}=\frac{10}{2t+1}-5.\)

At instantaneous rest, \(v=0\):

\(\frac{10}{2t+1}-5=0.\)

So \(2t+1=2\), giving

\(t=\frac12.\)

Now

\(s(0)=0,\)

\(s\left(\frac12\right)=5\ln2-2.5=0.9657\ldots,\)

and

\(s(2)=5\ln5-10=-1.9528\ldots.\)

The particle changes direction at \(t=\frac12\), so the distance travelled from \(0\) to \(2\) is

\(0.9657\ldots+(0.9657\ldots+1.9528\ldots)=3.88\text{ m}.\)

Acceleration is the derivative of velocity:

\(a=\frac{\mathrm dv}{\mathrm dt}=-\frac{20}{(2t+1)^2}.\)

When \(t=4.5\),

\(a=-\frac{20}{10^2}=-0.2\text{ m s}^{-2}.\)

This gives the required answer: \(t=\frac12\), distance \(=3.88\text{ m}\), acceleration \(=-\frac{20}{(2t+1)^2}\), and at \(t=4.5\), acceleration \(=-0.2\text{ m s}^{-2}\).