(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.