Answer: \(v=3(t-4)(t-2)\), \(a=6t-18\).

For \(s\): intercepts \((0,-16)\), \((1,0)\), \((4,0)\), with \(t=4\) a tangent point.

For \(v\): intercepts \((0,24)\), \((2,0)\), \((4,0)\). For \(a\): intercepts \((0,-18)\), \((3,0)\).

First expand the displacement:

\(s=(t-4)^2(t-1)=t^3-9t^2+24t-16.\)

The \(s\)-axis intercept is found by putting \(t=0\):

\(s=-16.\)

The \(t\)-axis intercepts are found from

\((t-4)^2(t-1)=0,\)

so \(t=1\) and \(t=4\). Since \(t=4\) is a repeated root, the displacement graph touches the \(t\)-axis there.

Differentiate to find the velocity:

\(v=\frac{\mathrm{d}s}{\mathrm{d}t}=3t^2-18t+24=3(t-4)(t-2).\)

So the velocity-time graph is an upward-opening parabola with \(t\)-intercepts \(2\) and \(4\), and \(v\)-intercept \(24\).

Differentiate again to find the acceleration:

\(a=\frac{\mathrm{d}v}{\mathrm{d}t}=6t-18.\)

The acceleration-time graph is a straight line with \(a\)-intercept \(-18\) and \(t\)-intercept \(3\).