\((i) To verify that P comes to rest at t = 200, substitute t = 200 into the velocity equation:\)

\(v(200) = 0.12 \times 200 - 0.0006 \times 200^2 = 0\).

\(Thus, P is at rest at t = 200. To find the acceleration, differentiate the velocity function:\)

\(a = \frac{dv}{dt} = 0.12 - 0.0012t\).

\(Substitute t = 200:\)

\(a(200) = 0.12 - 0.0012 \times 200 = -0.12 \text{ m/s}^2\).

Acceleration is 0.12 m/s² towards O.

(ii) To find the maximum speed, set \(\frac{dv}{dt} = 0\):

\(0.12 - 0.0012t = 0\).

Solve for t:

\(t = 100\).

Substitute back to find the maximum speed:

\(v(100) = 0.12 \times 100 - 0.0006 \times 100^2 = 6 \text{ m/s}\).

(iii) To find the displacement, integrate the velocity function:

\(s = \int (0.12t - 0.0006t^2) \, dt = 0.06t^2 - 0.0002t^3 + C\).

Assuming s = 0 when t = 0, \(C = 0\). Substitute t = 200:

\(s(200) = 0.06 \times 200^2 - 0.0002 \times 200^3 = 800 \text{ m}\).

(iv) To find when P reaches O again, set \(s = 0\):

\(0.06t^2 - 0.0002t^3 = 0\).

Factor out \(t^2\):

\(t^2(0.06 - 0.0002t) = 0\).

Thus, \(t = 0\) or \(0.06 = 0.0002t\).

Solve for t:

\(t = 300\).