(i) To find the acceleration, differentiate the velocity functions:

For \(0 \leq t \leq 5\), \(v(t) = 0.72t^2 - 0.096t^3\).

\(a_1(t) = \frac{d}{dt}(0.72t^2 - 0.096t^3) = 1.44t - 0.288t^2\).

For \(5 \leq t \leq 10\), \(v(t) = 2.4t - 0.24t^2\).

\(a_2(t) = \frac{d}{dt}(2.4t - 0.24t^2) = 2.4 - 0.48t\).

Evaluate at \(t = 5\):

\(a_1(5) = 1.44 \times 5 - 0.288 \times 25 = 0\).

\(a_2(5) = 2.4 - 0.48 \times 5 = 0\).

Since \(a_1(5) = a_2(5) = 0\), there is no instantaneous change in acceleration.

(ii) To find the distance \(AB\), integrate the velocity functions:

For \(0 \leq t \leq 5\), \(s_1 = \int_0^5 (0.72t^2 - 0.096t^3) \, dt = [0.24t^3 - 0.024t^4]_0^5\).

\(s_1 = (0.24 \times 5^3 - 0.024 \times 5^4) - (0 - 0) = 15\).

For \(5 \leq t \leq 10\), \(s_2 = \int_5^{10} (2.4t - 0.24t^2) \, dt = [1.2t^2 - 0.08t^3]_5^{10}\).

\(s_2 = (1.2 \times 10^2 - 0.08 \times 10^3) - (1.2 \times 5^2 - 0.08 \times 5^3) = 20\).

Total distance \(AB = s_1 + s_2 = 15 + 20 = 35\) m.

To find T₁ and T₂, we first find the acceleration function by differentiating the velocity function:

\(a(t) = \frac{d}{dt}(0.002t^3 - 0.12t^2 + 1.8t + 5) = 0.006t^2 - 0.24t + 1.8\)

Set \(a(t) = 0\) to find critical points:

\(0.006t^2 - 0.24t + 1.8 = 0\)

Solving the quadratic equation:

\(0.006(t^2 - 40t + 300) = 0\)

\(t^2 - 40t + 300 = 0\)

Using the quadratic formula:

\(t = \frac{40 \pm \sqrt{40^2 - 4 \times 1 \times 300}}{2 \times 1}\)

\(t = \frac{40 \pm \sqrt{1600 - 1200}}{2}\)

\(t = \frac{40 \pm 20}{2}\)

\(t = 30\) or \(t = 10\)

Thus, \(T_1 = 10\) and \(T_2 = 30\).

To find the distance \(OP\) when \(t = T_2\), integrate the velocity function:

\(s(t) = \int (0.002t^3 - 0.12t^2 + 1.8t + 5) \, dt\)

\(s(t) = 0.0005t^4 - 0.04t^3 + 0.9t^2 + 5t + C\)

Using initial condition \(s(0) = 0\), \(C = 0\).

Calculate \(s(T_2)\):

\(s(30) = 0.0005(30)^4 - 0.04(30)^3 + 0.9(30)^2 + 5(30)\)

\(s(30) = 405 - 1080 + 810 + 150\)

\(s(30) = 285\) m

For part (ii), the velocity at \(t = T_2\) is:

\(v(30) = 0.002(30)^3 - 0.12(30)^2 + 1.8(30) + 5\)

\(v(30) = 5\) m s^-1

The velocity-time graph shows \(v\) increasing from a positive value at \(t = 0\) to a maximum, then decreasing to a positive minimum, and thereafter increasing.

Given the velocity function \(v(t) = 1.2t - 0.12t^2\), we need to find the displacement when the acceleration is 0.6 m s^-2.

First, find the acceleration by differentiating the velocity function:

\(a(t) = \frac{dv}{dt} = 1.2 - 0.24t\)

Set the acceleration equal to 0.6:

\(1.2 - 0.24t = 0.6\)

Solve for \(t\):

\(0.24t = 0.6\)

\(t = 2.5\)

Now, find the displacement by integrating the velocity function from 0 to 2.5:

\(s = \int_0^{2.5} (1.2t - 0.12t^2) \, dt\)

\(s = \left[0.6t^2 - 0.04t^3\right]_0^{2.5}\)

\(s = (0.6 \times 2.5^2 - 0.04 \times 2.5^3) - (0 - 0)\)

\(s = 3.125\)

Therefore, the displacement is 3.125 m.

(i) To find \(A\), integrate the velocity function from \(t = 0\) to \(t = 15\):

\(\int_0^{15} v_1 \, dt = 225\)

\(\int_0^{15} A(t - 0.05t^2) \, dt = 225\)

\(A \left[ \frac{15^2}{2} - 0.05 \times \frac{15^3}{3} \right] = 225\)

\(A \left[ 112.5 - 0.05 \times 1125 \right] = 225\)

\(A \times 56.25 = 225\)

\(A = 4\)

To find \(B\), use the condition \(v_1(15) = v_2(15)\):

\(4(15 - 0.05 \times 15^2) = \frac{B}{15^2}\)

\(4 \times 11.25 = \frac{B}{225}\)

\(45 = \frac{B}{225}\)

\(B = 3375\)

(ii) For \(t \geq 15\), the distance travelled is:

\(s_2(t) = \int_{15}^{t} \frac{3375}{t^2} \, dt\)

\(= -3375 \left[ \frac{1}{t} - \frac{1}{15} \right]\)

\(= 225 - \frac{3375}{t}\)

Total distance: \(450 - \frac{3375}{t}\) m

(iii) Set the total distance to 315 m:

\(450 - \frac{3375}{t} = 315\)

\(\frac{3375}{t} = 135\)

\(t = 25\)

Speed at \(t = 25\):

\(v = \frac{3375}{25^2}\)

\(v = 5.4 \text{ m/s}\)

(i) The velocity is given by differentiating the displacement: \(v(t) = \frac{dx}{dt} = 1.2t - 0.012t^2\).

The acceleration is the derivative of velocity: \(a(t) = \frac{dv}{dt} = 1.2 - 0.024t\).

At \(t = 50\), \(a(50) = 1.2 - 0.024 \times 50 = 0\). Thus, the acceleration is zero.

The speed \(V\) at \(t = 50\) is \(v(50) = 1.2 \times 50 - 0.012 \times 50^2 = 30\) m s^-1.

(ii) For \(t \geq 50\), the motorcyclist travels at constant speed \(V = 30\) m s^-1.

The displacement at \(t = 50\) is \(s_1 = 0.6 \times 50^2 - 0.004 \times 50^3 = 1000\) m.

Let \(t_2\) be the time after \(t = 50\). The total time is \(t = 50 + t_2\).

The total distance is \(1000 + 30t_2\).

The average speed is given by \(\frac{1000 + 30t_2}{50 + t_2} = 27.5\).

Solving \(1000 + 30t_2 = 27.5(50 + t_2)\) gives \(t_2 = 150\).

Thus, \(t = 50 + 150 = 200\) seconds.