(i) The velocity-time graph consists of three straight line segments:

• The first segment has a positive slope, starting from 0 and reaching V at time t₁.
• The second segment is a horizontal line at velocity V from t₁ to t₁ + 600 s.
• The third segment has a negative slope, decreasing from V to 0 at time 1,000 s.

(ii) Using the area under the graph to represent the distance:

For the entire journey: \(\frac{1}{2} (600 + 1000) V = 20000\)

Solving for \(V\):

\(V = 25 \text{ m/s}\)

(iii) Given acceleration \(a = 0.15 \text{ m/s}^2\), find \(t_1\):

\(V = a t_1 \Rightarrow t_1 = \frac{V}{0.15} = \frac{25}{0.15} = 166.6 \text{ s}\)

Time for third stage \(t_3 = 1000 - (t_1 + 600) = 233.3 \text{ s}\)

Distance during third stage \(s_3 = \frac{1}{2} \times 233.3 \times 25 = 2920 \text{ m}\)