(i) To find \(k\), we need to determine when the velocity \(v(t) = k(60t^2 - t^3)\) is at its maximum. The derivative \(\frac{dv}{dt} = k(120t - 3t^2)\) is set to zero for maximum velocity:

\(120t - 3t^2 = 0\)

\(t(120 - 3t) = 0\)

\(t = 0 \text{ or } t = 40\)

At \(t = 40\), \(v(40) = k(60 \times 40^2 - 40^3) = 6.4\)

\(k(96000 - 64000) = 6.4\)

\(k \times 32000 = 6.4\)

\(k = \frac{6.4}{32000} = 0.0002\)

(ii) The particle comes to rest at \(t = 60\). Integrate \(v(t)\) to find \(s(t)\):

\(s(t) = \int k(60t^2 - t^3) \, dt = 0.0002 \left( 20t^3 - \frac{t^4}{4} \right) + C\)

Using \(t = 60\) and \(C = 0\):

\(OA = 0.0002 \left( 20 \times 60^3 - \frac{60^4}{4} \right)\)

\(OA = 0.0002 \times 1296000 = 216 \text{ m}\)

(iii) The acceleration at \(t = 60\) is given by:

\(\frac{dv}{dt} = 0.0002(120 \times 60 - 3 \times 60^2)\)

\(= 0.0002 \times 3600 = 0.72 \text{ m s}^{-2}\)

(iv) To find the speed when \(P\) passes through \(O\) again, solve \(s(t) = 0\) for non-zero \(t\):

\(20t^3 - \frac{t^4}{4} = 0\)

\(t^3(20 - \frac{t}{4}) = 0\)

\(t = 0 \text{ or } t = 80\)

At \(t = 80\), \(v(80) = 0.0002(60 \times 80^2 - 80^3)\)

\(= 0.0002(384000 - 512000)\)

\(= 0.0002 \times (-128000) = -25.6 \text{ m s}^{-1}\)

The speed is 25.6 m s^-1.