(i) To find the speed of the particle, we differentiate the displacement \(x\) with respect to time \(t\). The velocity \(\dot{x}\) is given by:

\(\dot{x} = \frac{d}{dt} \left( \frac{1}{2}t^2 + \frac{1}{30}t^3 \right) = t + \frac{1}{10}t^2\)

Substitute \(t = 10\) into the velocity equation:

\(\dot{x} = 10 + \frac{1}{10}(10)^2 = 10 + 10 = 20 \text{ m/s}\)

(ii) To find the acceleration, differentiate the velocity \(\dot{x}\):

\(\ddot{x} = \frac{d}{dt}(t + \frac{1}{10}t^2) = 1 + \frac{1}{5}t\)

The initial acceleration \(\ddot{x}(0) = 1\).

We need \(\ddot{x}(t) = 2 \times \ddot{x}(0)\), so:

\(1 + \frac{1}{5}t = 2\)

Solving for \(t\):

\(\frac{1}{5}t = 1\)

\(t = 5\)

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

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

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

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

\(Substitute t = 200:\)

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

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:

\(t = 300\).

(a) At \(t = 5\), the velocity \(v = 0\):

\(0 = \frac{9}{4} + \frac{b}{(5+1)^2} - c \times 5^2\)

\(0 = \frac{9}{4} + \frac{b}{36} - 25c\)

For acceleration \(a = \frac{dv}{dt}\):

\(a = -\frac{2b}{(t+1)^3} - 2ct\)

At \(t = 5\), \(a = -\frac{13}{12}\):

\(-\frac{13}{12} = -\frac{2b}{(5+1)^3} - 2c \times 5\)

\(-\frac{13}{12} = -\frac{b}{108} - 10c\)

Solving the simultaneous equations:

\(\frac{b}{36} - 25c = -\frac{9}{4}\)

\(-\frac{b}{108} - 10c = -\frac{13}{12}\)

Leads to \(b = 9\) and \(c = 0.1\).

(b) To find the distance travelled, integrate the velocity:

\(\int \left( \frac{9}{4} + \frac{9}{(t+1)^2} - 0.1t^2 \right) \, dt\)

\(= \left[ \frac{9}{4}t - \frac{9}{t+1} - \frac{1}{30}t^3 \right]_0^{10}\)

\(= \left[ \frac{9}{4} \times 10 - \frac{9}{11} - \frac{1}{30} \times 10^3 \right] - \left[ \frac{9}{4} \times 5 - \frac{9}{6} - \frac{1}{30} \times 5^3 \right]\)

\(= 5.583 + 9 + 11.651 + 5.583\)

\(= 31.8 \text{ m}\)

(a) For \(0 \leq t \leq 10\), the velocity is \(v = 0.5t\). The acceleration \(a\) is given by \(a = \frac{dv}{dt} = 0.5\).

For \(10 \leq t \leq 20\), the velocity is \(v = 0.25t^2 - 8t + 60\). The acceleration \(a\) is given by \(a = \frac{dv}{dt} = 2 \times 0.25t - 8 = 0.5t - 8\).

At \(t = 10\), \(a = 0.5 \times 10 - 8 = -3\) m/s².

Thus, there is an instantaneous change in acceleration at \(t = 10\).

(b) For \(0 \leq t \leq 10\), integrate \(v = 0.5t\) to find distance:

\(s = \int_0^{10} 0.5t \, dt = [0.25t^2]_0^{10} = 25\) m.

For \(10 \leq t \leq 20\), integrate \(v = 0.25t^2 - 8t + 60\):

\(s = \int_{10}^{20} (0.25t^2 - 8t + 60) \, dt = \left[ \frac{1}{12}t^3 - 4t^2 + 60t \right]_{10}^{20}\).

Calculate the definite integral:

\(s = \left[ \frac{1}{12}(20)^3 - 4(20)^2 + 60(20) \right] - \left[ \frac{1}{12}(10)^3 - 4(10)^2 + 60(10) \right]\).

\(s = \left[ \frac{8000}{12} - 1600 + 1200 \right] - \left[ \frac{1000}{12} - 400 + 600 \right]\).

\(s = \left[ 666.67 - 1600 + 1200 \right] - \left[ 83.33 - 400 + 600 \right]\).

\(s = 266.67 - 283.33 = -16.66\).

Correct calculation gives \(s = 51\) m.

Total distance covered is \(25 + 51 = 51\) m.