(i) To find the maximum velocity during the first 2 seconds, differentiate \(v = 12t - 4t^2\) with respect to \(t\):

\(\frac{dv}{dt} = 12 - 8t\).

Set \(\frac{dv}{dt} = 0\) to find the critical points:

\(12 - 8t = 0 \Rightarrow t = 1.5\).

Substitute \(t = 1.5\) into the velocity equation:

\(v = 12 \times 1.5 - 4 \times (1.5)^2 = 9\) m s^-1.

(ii) For \(0 \leq t \leq 2\), the acceleration is \(\frac{dv}{dt} = 12 - 8t\). At \(t = 2\), \(\frac{dv}{dt} = 12 - 8 \times 2 = -4\).

For \(2 \leq t \leq 4\), the acceleration is \(\frac{dv}{dt} = -4\).

Since the acceleration is \(-4\) in both cases, there is no instantaneous change at \(t = 2\).

(iii) The velocity-time graph is a quadratic curve for \(0 \leq t \leq 2\) with a maximum at \(t = 1.5\), and a straight line with a negative gradient from \(t = 2\) to \(t = 4\).

(iv) The distance travelled is the area under the velocity-time graph. For \(0 \leq t \leq 2\), integrate \(12t - 4t^2\):

\(\int (12t - 4t^2) \, dt = 6t^2 - \frac{4}{3}t^3\) from 0 to 2.

\(= 6 \times 2^2 - \frac{4}{3} \times 2^3 = 16\).

For \(2 \leq t \leq 4\), the area of the triangle is \(\frac{1}{2} \times 2 \times 8 = 8\).

\(Total distance = 16 + 8 = 24.\)

Distance travelled is \(\frac{64}{3}\) m or 21.3 m.