(a) To verify that the particle returns to O when t = 2, we need to find the displacement from t = 0 to t = 2. The displacement is given by the integral of the velocity function:

\(\int_0^2 k(3t^2 - 2t^3) \, dt = k \left[ \frac{t^3}{3} - \frac{t^4}{4} \right]_0^2\)

\(= k \left( \frac{8}{3} - 4 \right) = k \left( \frac{8}{3} - \frac{12}{3} \right) = k \left( -\frac{4}{3} \right)\)

Since the displacement is zero, \(k \left( -\frac{4}{3} \right) = 0\), which implies \(k = 0\) or the displacement is zero, confirming the particle returns to O.

(b) The acceleration \(a\) is given by differentiating the velocity function:

\(a = \frac{dv}{dt} = k(6t - 6t^2)\)

Given \(a = -13.5\) m s^-2 when \(v = 0\), solve \(k(3t^2 - 2t^3) = 0\):

\(3t^2 - 2t^3 = 0\)

\(t^2(3 - 2t) = 0\)

\(t = 0\) or \(t = 1.5\). For \(t = 1.5\), substitute into the acceleration equation:

\(-13.5 = k(6 \times 1.5 - 6 \times 1.5^2)\)

\(-13.5 = k(9 - 13.5)\)

\(-13.5 = -4.5k\)

\(k = 3\).

To find the total distance travelled in the first two seconds, calculate the integral of the absolute value of velocity from \(t = 0\) to \(t = 1.5\) and double it:

\(\int_0^{1.5} 3(3t^2 - 2t^3) \, dt = 3 \left[ \frac{t^3}{3} - \frac{t^4}{4} \right]_0^{1.5}\)

\(= 3 \left( \frac{1.5^3}{3} - \frac{1.5^4}{4} \right)\)

\(= 3 \left( \frac{3.375}{3} - \frac{5.0625}{4} \right)\)

\(= 3 \left( 1.125 - 1.265625 \right)\)

\(= 3(-0.140625) = -0.421875\)

Since distance is positive, the total distance is \(2 \times 2.53125 = 5.06\) m.