(i) To find the distance \(OP\), integrate the velocity function:

\(v = 0.6t^2 - 0.12t^3\).

Set \(v = 0\) to find when the particle comes to rest:

\(0.6t^2 - 0.12t^3 = 0\)

\(t^2(0.6 - 0.12t) = 0\)

\(t = 0\) or \(t = 5\).

Integrate \(v\) to find \(s\):

\(s = \int (0.6t^2 - 0.12t^3) \, dt = 0.2t^3 - 0.03t^4 + C\).

Since the particle starts from rest at \(O\), \(C = 0\).

Evaluate from \(t = 0\) to \(t = 5\):

\(OP = [0.2 \times 5^3 - 0.03 \times 5^4] - [0] = 6.25\) m.

(ii) Given \(s = kt^3 + ct^5\) and \(s = 6.25\) at \(t = 5\):

\(k \times 5^3 + c \times 5^5 = 6.25\).

Differentiate \(s\) to find \(v\):

\(v = 3kt^2 + 5ct^4\).

Given \(v = 1.25\) at \(t = 5\):

\(1.25 = 3k \times 5^2 + 5c \times 5^4\).

Solve the simultaneous equations:

\(125k + 3125c = 6.25\)

\(75k + 3125c = 1.25\).

Solving gives \(k = 0.1, c = -0.002\).

(iii) Differentiate \(v = 0.6t^2 - 0.12t^3\) to find acceleration \(a\):

\(a = \frac{dv}{dt} = 1.2t - 0.36t^2\).

At \(t = 5\), \(a = 1.2 \times 5 - 0.36 \times 5^2 = -2\) m s^-2.