Answer: (a) \(t=\dfrac{\pi}{2}\); (b) \(s=\dfrac32-\dfrac32\cos2t\); (c) \(6\).

We start with the main method. Displacement is found by integrating velocity over the given time interval.

The velocity is

\(v=3\sin2t.\)

For the velocity to be zero,

\(\sin2t=0.\)

The first positive value after \(t=0\) is

\(2t=\pi\), so

\(t=\frac{\pi}{2}.\)

For part (b), displacement is the integral of velocity:

\(s=\int 3\sin2t\,dt.\)

So

\(s=-\frac32\cos2t+C.\)

At \(t=0\), the particle is at \(O\), so \(s=0\).

Therefore

\(0=-\frac32\cos0+C=-\frac32+C,\)

giving \(C=\frac32\).

Hence

\(s=\frac32-\frac32\cos2t.\)

For \(0\leq t\leq \pi\), the velocity is positive on \(0\lt t\lt \frac{\pi}{2}\) and negative on \(\frac{\pi}{2}\lt t\lt \pi\).

The displacement at \(t=\frac{\pi}{2}\) is

\(s=\frac32-\frac32\cos\pi=3.\)

The displacement at \(t=\pi\) is

\(s=\frac32-\frac32\cos2\pi=0.\)

So the distance travelled is

\(3+3=6.\)

This completes the solution and gives the required result.