(i) To find k, integrate the acceleration to get the velocity:

\(v = \int k(3t^2 - 12t + 2) \, dt = k \left( \frac{3t^3}{3} - \frac{12t^2}{2} + 2t \right) + C\)

\(v = k(t^3 - 6t^2 + 2t) + C\)

Given initial velocity \(v = 0.4\) m/s when \(t = 0\), \(C = 0.4\).

At \(t = 1\), \(v = 0.1\):

\(0.1 = k(1 - 6 + 2) + 0.4\)

\(0.1 = k(-3) + 0.4\)

\(-0.3 = -3k\)

\(k = 0.1\)

(ii) Integrate the velocity to find displacement:

\(s = \int (0.1(t^3 - 6t^2 + 2t) + 0.4) \, dt\)

\(s = 0.1 \left( \frac{t^4}{4} - \frac{6t^3}{3} + \frac{2t^2}{2} \right) + 0.4t + C\)

\(s = 0.025t^4 - 0.2t^3 + 0.1t^2 + 0.4t\)

Since the particle starts from the origin, \(C = 0\).

(iii) Verify at \(t = 2\):

\(s = 0.025(2)^4 - 0.2(2)^3 + 0.1(2)^2 + 0.4(2)\)

\(s = 0.4 - 1.6 + 0.4 + 0.8 = 0\)

The particle is at the origin at \(t = 2\).