Answer: distance \(4\text{ m}\); acceleration \(-18\text{ m s}^{-2}\).

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

\(3t^2-30t+72=0.\)

Divide by \(3\):

\(t^2-10t+24=0.\)

Factorise:

\((t-4)(t-6)=0.\)

So the particle is at rest at \(t=4\) and \(t=6\).

Between \(t=4\) and \(t=6\), the velocity is negative, so the distance is

\(-\int_4^6 (3t^2-30t+72)\,dt.\)

An antiderivative is

\(t^3-15t^2+72t.\)

Thus the displacement from \(t=4\) to \(t=6\) is

\([t^3-15t^2+72t]_4^6=108-112=-4.\)

Therefore the distance is

\(4\text{ m}.\)

(b) Acceleration is

\(a=\frac{dv}{dt}=6t-30.\)

When \(t=2\),

\(a=6(2)-30=-18.\)

So the acceleration is

\(-18\text{ m s}^{-2}.\)