(i) To find when P returns to O, set \(x = 0\):

\(0.08t^2 - 0.0002t^3 = 0\)

\(t^2(0.08 - 0.0002t) = 0\)

\(t = 0\) or \(0.08 - 0.0002t = 0\)

\(0.0002t = 0.08\)

\(t = \frac{0.08}{0.0002} = 400\) s

To find the speed at O on return, differentiate \(x\) to get velocity \(v\):

\(v = \frac{dx}{dt} = 0.16t - 0.0006t^2\)

At \(t = 400\):

\(v = 0.16 \times 400 - 0.0006 \times 400^2\)

\(v = 64 - 96 = -32\) m/s

Speed is 32 m/s.

(ii)(a) To find the total distance travelled, find the time to the furthest point:

\(v = 0.16t - 0.0006t^2 = 0\)

\(t(0.16 - 0.0006t) = 0\)

\(t = 0\) or \(t = \frac{0.16}{0.0006} \approx 266.67\) s

Distance to furthest point:

\(x = 0.08(266.67)^2 - 0.0002(266.67)^3\)

\(x \approx 1895\) m

Total distance = \(2 \times 1895 = 3790\) m

(ii)(b) Average speed = \(\frac{\text{total distance}}{\text{total time}} = \frac{3790}{400} = 9.48\) m/s

(i) To find the acceleration, differentiate the velocity function \(v(t) = 0.00004t^3 - 0.006t^2 + 0.288t\) with respect to \(t\):

\(a(t) = \frac{dv}{dt} = 0.00012t^2 - 0.012t + 0.288\).

Set \(a(t) = 0\) to find when the acceleration is zero:

\(0.00012t^2 - 0.012t + 0.288 = 0\).

Factor the quadratic equation:

\(0.00012(t^2 - 100t + 2400) = 0.00012(t - 40)(t - 60) = 0\).

Thus, \(t = 40\) and \(t = 60\).

(ii) To find the displacement, integrate the velocity function from \(t = 0\) to \(t = 100\):

\(\int_0^{100} (0.00004t^3 - 0.006t^2 + 0.288t) \, dt\).

Calculate the integral:

\(\left[ 0.00001t^4 - 0.002t^3 + 0.144t^2 \right]_0^{100}\).

Evaluate at the limits:

\(0.00001(100)^4 - 0.002(100)^3 + 0.144(100)^2 = 440\).

Therefore, the displacement is 440 m.

(i) To find the displacement \(s\), integrate the velocity function:

\(s = \int (0.6t - 0.03t^2) \, dt = 0.3t^2 - 0.01t^3 + C\).

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

Calculate \(s(5)\):

\(s(5) = 0.3 \times 5^2 - 0.01 \times 5^3 = 6.25\).

To find the acceleration, differentiate the velocity function:

\(a = \frac{dv}{dt} = 0.6 - 0.06t\).

Calculate \(a(5)\):

\(a(5) = 0.6 - 0.06 \times 5 = 0.3 \text{ m s}^{-2}\).

(ii) Maximum velocity occurs when \(a = 0\):

\(0.6 - 0.06t = 0 \Rightarrow t = 10\).

Maximum velocity is \(v(10) = 3 \text{ m s}^{-1}\).

Half of maximum velocity is \(1.5 \text{ m s}^{-1}\).

Set \(0.6t - 0.03t^2 = 1.5\):

\(0.03t^2 - 0.6t + 1.5 = 0\).

Solving the quadratic equation:

\(t^2 - 20t + 50 = 0\).

Using the quadratic formula, \(t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), find:

\(t = 2.93 \text{ s}\) and \(t = 17.07 \text{ s}\).

To find the displacement, we first integrate the acceleration to find the velocity. Given:

\(a(t) = 0.075t^2 - 1.5t + 5\)

Integrate \(a(t)\) to find \(v(t)\):

\(v(t) = \int (0.075t^2 - 1.5t + 5) \, dt = 0.025t^3 - 0.75t^2 + 5t + C\)

Since the particle starts from rest, \(v(0) = 0\), so \(C = 0\).

Thus, \(v(t) = 0.025t^3 - 0.75t^2 + 5t\).

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

\(s(t) = \int (0.025t^3 - 0.75t^2 + 5t) \, dt = 0.00625t^4 - 0.25t^3 + 2.5t^2 + C\)

Since the displacement is zero at \(t = 0\), \(C = 0\).

Thus, \(s(t) = 0.00625t^4 - 0.25t^3 + 2.5t^2\).

To find when \(s(t) = 0\) again, solve:

\(0.00625t^4 - 0.25t^3 + 2.5t^2 = 0\)

Factor out \(t^2\):

\(t^2(0.00625t^2 - 0.25t + 2.5) = 0\)

\(t^2 = 0\) gives \(t = 0\), and solving \(0.00625t^2 - 0.25t + 2.5 = 0\) gives \(t = 20\).

Therefore, the time taken for P to return to the point O is 20 seconds.

(i) For the first 8 seconds, the velocity \(v(8) = 0.25 \times 8 = 2\) m/s. Using the given velocity equation:

\(v = -0.1t^2 + 2.4t - k\)

Substitute \(t = 8\):

\(2 = -0.1(8)^2 + 2.4(8) - k\)

\(2 = -6.4 + 19.2 - k\)

\(k = 10.8\)

(ii) To find the maximum velocity, differentiate \(v\) with respect to \(t\) and set the derivative to zero:

\(\frac{dv}{dt} = -0.2t + 2.4\)

Set \(\frac{dv}{dt} = 0\):

\(-0.2t + 2.4 = 0\)

\(t = 12\)

Substitute \(t = 12\) into the velocity equation:

\(v = -0.1(12)^2 + 2.4(12) - 10.8\)

\(v = -14.4 + 28.8 - 10.8\)

\(v = 3.6\) m/s

(iii) Displacement for the first 8 seconds:

\(s_1 = \frac{1}{2} \times 0.25 \times 8^2 = 8\) m

Displacement from \(t = 8\) to \(t = 18\):

\(s_2 = \int_{8}^{18} (-0.1t^2 + 2.4t - 10.8) \, dt\)

\(s_2 = \left[ -0.1 \frac{t^3}{3} + 1.2t^2 - 10.8t \right]_{8}^{18}\)

\(s_2 = [-0.1(18)^3/3 + 1.2(18)^2 - 10.8(18)] - [-0.1(8)^3/3 + 1.2(8)^2 - 10.8(8)]\)

\(s_2 = 26.7\) m

Total displacement \(s = s_1 + s_2 = 8 + 26.7 = 34.7\) m