(i) The acceleration is given by \(a(t) = 0.6t\). Integrating to find velocity:

\(v(t) = \int 0.6t \, dt = 0.3t^2 + C\).

Since the particle starts from rest, \(v(0) = 0\), so \(C = 0\). Thus, \(v(t) = 0.3t^2\).

At \(t = 10\), \(v(10) = 0.3 \times 10^2 = 30\) m s⁻¹.

Integrating velocity to find displacement:

\(s(t) = \int 0.3t^2 \, dt = 0.1t^3 + C\).

Since \(s(0) = 0\), \(C = 0\). Thus, \(s(t) = 0.1t^3\).

At \(t = 10\), \(s(10) = 0.1 \times 10^3 = 100\) m.

(ii) After \(t = 10\), the acceleration is \(a(t) = -0.4t\). Integrating to find velocity:

\(v(t) = \int -0.4t \, dt = -0.2t^2 + C\).

Using \(v(10) = 30\), \(-0.2 \times 10^2 + C = 30\) gives \(C = 50\). Thus, \(v(t) = -0.2t^2 + 50\).

At point \(A\), \(v(t) = 0\):

\(-0.2t^2 + 50 = 0\) gives \(t = \sqrt{250}\).

Integrating velocity to find displacement:

\(s(t) = \int (-0.2t^2 + 50) \, dt = -\frac{t^3}{15} + 50t + C\).

Using \(s(10) = 100\), \(-\frac{10^3}{15} + 500 + C = 100\) gives \(C = -\frac{1000}{3}\).

Thus, \(s(t) = -\frac{t^3}{15} + 50t - \frac{1000}{3}\).

At \(t = \sqrt{250}\), \(s(\sqrt{250}) = -\frac{(\sqrt{250})^3}{15} + 50\sqrt{250} - \frac{1000}{3} = 194\) m.

(i) To find the time t when the aeroplane takes off, we need to find the velocity function by integrating the acceleration function. The acceleration is given by:

\(a(t) = 1.5 + 0.012t\)

Integrating with respect to t gives the velocity function:

\(v(t) = \int (1.5 + 0.012t) \, dt = 1.5t + 0.006t^2 + C\)

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

Thus, the velocity function is:

\(v(t) = 1.5t + 0.006t^2\)

We set \(v(t) = 90\) to find the time when the aeroplane takes off:

\(1.5t + 0.006t^2 = 90\)

Solving the quadratic equation:

\(0.006t^2 + 1.5t - 90 = 0\)

\(t^2 + 250t - 15000 = 0\)

\((t - 50)(t + 300) = 0\)

Thus, \(t = 50\) (since time cannot be negative).

(ii) To find the distance travelled, integrate the velocity function:

\(s(t) = \int (1.5t + 0.006t^2) \, dt = 0.75t^2 + 0.002t^3 + C\)

Since \(s(0) = 0\), \(C = 0\).

Thus, the distance function is:

\(s(t) = 0.75t^2 + 0.002t^3\)

Substitute \(t = 50\) to find the distance:

\(s(50) = 0.75(50)^2 + 0.002(50)^3\)

\(= 0.75(2500) + 0.002(125000)\)

\(= 1875 + 250\)

\(= 2125\) meters.

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

\(a(t) = \frac{dv}{dt} = \frac{d}{dt}(0.2t + 0.006t^2) = 0.2 + 0.012t\).

The initial acceleration is \(a(0) = 0.2\).

We are given that \(a(t) = 2.5 \times a(0)\), so:

\(0.2 + 0.012t = 2.5 \times 0.2\).

Simplifying gives:

\(0.2 + 0.012t = 0.5\).

\(0.012t = 0.3\).

\(t = \frac{0.3}{0.012} = 25\).

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

\(s(t) = \int (0.2t + 0.006t^2) \, dt = 0.1t^2 + 0.002t^3 + C\).

Assuming \(C = 0\) (since the particle starts from O), evaluate from \(t = 0\) to \(t = 25\):

\(s(25) = 0.1 \times 25^2 + 0.002 \times 25^3\).

\(s(25) = 0.1 \times 625 + 0.002 \times 15625\).

\(s(25) = 62.5 + 31.25 = 93.75\).

(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.

Using the equation of motion \(s = ut + \frac{1}{2}at^2\) for the segment \(AB\):

\(55 = 5u + \frac{1}{2}a(5^2)\)

\(55 = 5u + 12.5a\) (Equation 1)

For the segment \(BC\), the final velocity at \(B\) is \(v_B = u + 5a\). Using the equation \(s = ut + \frac{1}{2}at^2\) again:

\(65 = 5v_B + \frac{1}{2}a(5^2)\)

\(65 = 5(u + 5a) + 12.5a\)

\(65 = 5u + 25a + 12.5a\)

\(65 = 5u + 37.5a\) (Equation 2)

Subtract Equation 1 from Equation 2:

\(65 - 55 = (5u + 37.5a) - (5u + 12.5a)\)

\(10 = 25a\)

\(a = 0.4 \text{ m s}^{-2}\)

Substitute \(a = 0.4\) into Equation 1:

\(55 = 5u + 12.5(0.4)\)

\(55 = 5u + 5\)

\(50 = 5u\)

\(u = 10 \text{ m s}^{-1}\)