Answer: (a) \(2+\frac{7\pi}{6}+\sqrt3\); (b) \(\frac{11\pi}{6}\); (c) \(2+\frac{5\pi}{6}+2\sqrt3\).

Differentiate displacement to get velocity:

\(v=\frac{ds}{dt}=1+2\sin t.\)

(a) Instantaneous rest occurs when \(v=0\):

\(1+2\sin t=0.\)

So

\(\sin t=-\frac12.\)

For \(t\gt 0\), the first such time is

\(t=\frac{7\pi}{6}.\)

The displacement then is

\(s=2+\frac{7\pi}{6}-2\cos\frac{7\pi}{6}.\)

Since \(\cos\frac{7\pi}{6}=-\frac{\sqrt3}{2}\),

\(s=2+\frac{7\pi}{6}+\sqrt3.\)

(b) The next time when \(\sin t=-\frac12\) is

\(t=\frac{11\pi}{6}.\)

(c) For \(0\leq t\leq\frac{3\pi}{2}\), the particle moves forwards until \(t=\frac{7\pi}{6}\), then backwards until \(t=\frac{3\pi}{2}\).

Now

\(s(0)=2+0-2=0,\)

and

Also

The total distance travelled is

So

Therefore

\(\text{distance}=2+\frac{5\pi}{6}+2\sqrt3.\)