Answer: \(v=4+8\sin2t\); \(t\)-intercepts \(t=\dfrac{7\pi}{12},\dfrac{11\pi}{12}\); \(a=16\cos2t\); \(a=0\) when \(t=\dfrac{\pi}{4},\dfrac{3\pi}{4}\).

We start with the main method. Use \(v=ds/dt\) and \(a=dv/dt\), then substitute the requested value of time.

The displacement is

\(s=4t-4\cos2t+4.\)

Velocity is the derivative of displacement:

\(v=\frac{ds}{dt}.\)

Thus

\(v=4+8\sin2t.\)

For the velocity-time graph, when \(t=0\),

\(v=4+8\sin0=4.\)

So the graph crosses the \(v\)-axis at \(v=4\).

For the \(t\)-intercepts, set \(v=0\):

\(4+8\sin2t=0.\)

So

\(\sin2t=-\frac12.\)

Since \(0\leqslant t\leqslant\pi\), we have \(0\leqslant2t\leqslant2\pi\).

In this interval,

\(2t=\frac{7\pi}{6}\quad\text{or}\quad2t=\frac{11\pi}{6}.\)

Hence

\(t=\frac{7\pi}{12}\quad\text{or}\quad t=\frac{11\pi}{12}.\)

The sketch is one complete sine cycle shifted upwards by \(4\), with amplitude \(8\), from \(t=0\) to \(t=\pi\).

For part (c), acceleration is the derivative of velocity:

\(a=\frac{dv}{dt}=16\cos2t.\)

For part (d), set acceleration equal to zero:

\(16\cos2t=0.\)

So

\(\cos2t=0.\)

For \(0\leqslant2t\leqslant2\pi\),

\(2t=\frac{\pi}{2}\quad\text{or}\quad2t=\frac{3\pi}{2}.\)

Therefore

\(t=\frac{\pi}{4}\quad\text{or}\quad t=\frac{3\pi}{4}.\)

This completes the solution and gives the required result.