Answer: (a) \(\displaystyle t=\frac{2\pi}{9}\); (b) displacement \(=2.92\text{ m}\).

Answer: (a) \(\displaystyle t=\frac{2\pi}{9}\); (b) displacement \(=2.92\text{ m}\).

(a) Acceleration is the derivative of velocity:

\(a=\frac{dv}{dt}=18\cos3t.\)

We need

\(18\cos3t=-9.\)

So

\(\cos3t=-\frac12.\)

The first positive solution is

\(3t=\frac{2\pi}{3}.\)

Hence

\(t=\frac{2\pi}{9}.\)

(b) Displacement is found by integrating velocity:

\(s=\int6\sin3t\,dt=-2\cos3t+C.\)

The particle passes through \(O\) when \(t=0\), so \(s=0\) at \(t=0\). Therefore

\(0=-2\cos0+C=-2+C,\)

so \(C=2\). Hence

\(s=2-2\cos3t.\)

When \(t=5.6\),

\(s=2-2\cos(16.8)=2.92\ldots.\)

Therefore the displacement is \(2.92\text{ m}\) to \(3\) significant figures.