(i) The velocity-time graph consists of three segments: an upward sloping line from 0 to V over T₁ seconds, a horizontal line at V for T₂ seconds, and a downward sloping line from V to 0 over T₃ seconds. Using the equations of motion:

For acceleration: \(v = u + at \Rightarrow V = 0 + 0.3T_1 \Rightarrow T_1 = \frac{V}{0.3}\).

For deceleration: \(v = u - at \Rightarrow 0 = V - 1T_3 \Rightarrow T_3 = V\).

(ii) The total distance travelled is the area under the velocity-time graph, which is a trapezium:

\(S = \frac{1}{2} T_1 V + T_2 V + \frac{1}{2} T_3 V\).

Substitute \(T_1 = \frac{V}{0.3}\) and \(T_3 = V\):

\(S = \frac{1}{2} \left(\frac{V}{0.3}\right) V + T_2 V + \frac{1}{2} V^2\).

Given \(S = 12000\) m and \(T_1 + T_2 + T_3 = 552\):

\(12000 = \frac{1}{2} \left(\frac{V}{0.3}\right) V + T_2 V + \frac{1}{2} V^2\).

\(12000 = \frac{V^2}{0.6} + T_2 V + \frac{V^2}{2}\).

\(12000 = \frac{5V^2}{6} + T_2 V\).

\(T_2 = 552 - \frac{V}{0.3} - V\).

Substitute \(T_2\) into the equation:

\(12000 = \frac{5V^2}{6} + (552 - \frac{V}{0.3} - V)V\).

\(12000 = \frac{5V^2}{6} + 552V - \frac{V^2}{0.3} - V^2\).

\(12000 = \frac{5V^2}{6} + 552V - \frac{10V^2}{3} - V^2\).

\(12000 = 552V - \frac{13V^2}{6}\).

Multiply through by 6 to clear the fraction:

\(72000 = 3312V - 13V^2\).

Rearrange to form a quadratic equation:

\(13V^2 - 3312V + 72000 = 0\).

Solving this quadratic equation gives \(V = 24\).