Answer: \(v=9t(t-4)\), so \(v=0\) at \(t=0\) and \(t=4\). The acceleration is \(18t-36\).

Use the given velocity, acceleration or displacement relationship consistently. Remember that acceleration is the derivative of velocity, and displacement is obtained by integrating velocity.

(a) The displacement is

\(s=3(t+2)(t-4)^2.\)

Differentiate to find the velocity:

\(v=\frac{\mathrm ds}{\mathrm dt} =3(t-4)^2+3(t+2)\cdot2(t-4).\)

Factorise:

\(v=3(t-4)\bigl((t-4)+2(t+2)\bigr).\)

So

\(v=3(t-4)(3t)=9t(t-4).\)

Thus \(v=0\) when

\(t=0 \quad\text{or}\quad t=4.\)

(b) The displacement-time graph is

\(s=3(t+2)(t-4)^2.\)

The \(s\)-axis intercept is

\(s(0)=3(2)(16)=96.\)

The \(t\)-axis intercept in the interval is

\(t=4.\)

At \(t=4\), the graph touches the \(t\)-axis because \((t-4)^2\) is a repeated factor. The graph decreases from \((0,96)\) to \((4,0)\), then increases again for \(4\lt t\leq5\).

(c) The velocity-time graph is

\(v=9t(t-4).\)

This is an upward-opening parabola with intercepts

\((0,0) \quad\text{and}\quad (4,0).\)

The vertex occurs halfway between the roots, at \(t=2\). Then

\(v(2)=9(2)(-2)=-36.\)

So the velocity graph has minimum point \((2,-36)\).

(d)(i) The acceleration is

\(\frac{\mathrm dv}{\mathrm dt}.\)

Since

\(v=9t^2-36t,\)

we get

\(\text{acceleration}=18t-36.\)

(d)(ii) The acceleration-time graph is the straight line

\(a=18t-36.\)

It meets the acceleration axis at

\((0,-36),\)

and the \(t\)-axis when

\(18t-36=0,\)

so

\(t=2.\)

Thus the intercepts are \((0,-36)\) and \((2,0)\).

Therefore the result matches the required answer.