Answer: \(t=0.615\text{ s}\); displacement \(=\dfrac32-\dfrac{\pi}{4}\); acceleration \(=-4\sqrt2\text{ m s}^{-2}\).

The particle is at rest when \(v=0\):

\(3\cos 2t-1=0.\)

So

\(\cos 2t=\frac13.\)

The first positive value is

\(t=\frac12\cos^{-1}\left(\frac13\right)=0.615\text{ s}\quad \text{to 3 significant figures}.\)

Displacement is found by integrating velocity:

\(s=\int(3\cos2t-1)\,dt=\frac32\sin2t-t+C.\)

Since the particle passes through \(O\) when \(t=0\), \(s=0\) at \(t=0\), so \(C=0\).

At \(t=\dfrac{\pi}{4}\),

\(s=\frac32\sin\frac{\pi}{2}-\frac{\pi}{4} =\frac32-\frac{\pi}{4}.\)

The acceleration is

\(a=\frac{dv}{dt}=-6\sin2t.\)

At the first instant of rest, \(\cos2t=\dfrac13\) and \(2t\) is acute, so

\(\sin2t=\sqrt{1-\frac19}=\frac{2\sqrt2}{3}.\)

Therefore

\(a=-6\cdot\frac{2\sqrt2}{3}=-4\sqrt2.\)