(i) The velocity-time graph is a trapezium with three straight lines: a positive gradient for the first 3 seconds, a zero gradient for the next 6 seconds, and a negative gradient for the last 4 seconds. The time points 0, 3, 9, and 13 are shown on the time axis. The maximum velocity reached is 2.7 m/s.

(ii) To find the total distance, we calculate the area under the velocity-time graph, which is a trapezium. The formula for the area of a trapezium is:

\(\text{Area} = \frac{1}{2} \times (\text{Base}_1 + \text{Base}_2) \times \text{Height}\)

Here, the bases are the time intervals: 6 s and 13 s, and the height is the maximum velocity 2.7 m/s. Thus, the total distance is:

\(\text{Total distance} = \frac{1}{2} \times (6 + 13) \times 2.7 = 25.65 \text{ m}\)

Alternatively, using the constant acceleration equations for each stage:

1. Stage 1: \(s_1 = 0.5 \times 0.9 \times 3^2 = 4.05 \text{ m}\)
2. Stage 2: \(s_2 = 2.7 \times 6 = 16.2 \text{ m}\)
3. Stage 3: \(s_3 = 0.5 \times (2.7 + 0) \times 4 = 5.4 \text{ m}\)

Total distance = \(s_1 + s_2 + s_3 = 4.05 + 16.2 + 5.4 = 25.65 \text{ m}\)

(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 \(V = 24\).

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

• From \(t = 0\) to \(t = 8\), velocity increases from 0 to 8 m/s.
• From \(t = 8\) to \(t = 20\), velocity decreases from 8 m/s to 2 m/s.
• From \(t = 20\) to \(t = 26\), velocity increases from 2 m/s to 6.5 m/s.

(ii) The shaded region for \(20 \leq t \leq 26\) is the area under the velocity-time graph between these times.

(iii) To show \(s = 0.375t^2 - 13t + 202\):

1. Calculate \(s(20)\):

\(s(20) = \frac{1}{2} \times 8 \times 8 + \frac{1}{2} \times (8 + 2) \times 12 = 92 \text{ meters}.\)

2. Calculate acceleration in the third phase:

\(a = \frac{6.5 - 2}{6} = 0.75 \text{ m/s}^2.\)

3. Use the equation for displacement:

\(s(t) = 92 + 2(t - 20) + 0.375(t - 20)^2.\)

4. Simplify to get:

\(s(t) = 0.375t^2 - 13t + 202.\)

(i) The train accelerates from rest with an acceleration of 0.025 m s^-2 for 600 s. The velocity at the end of this period is given by:

\(v(600) = 0.025 \times 600 = 15 \text{ m s}^{-1}\)

The train then travels at a constant speed of 15 m s^-1 for 2600 s. Finally, it decelerates to rest with a deceleration of 0.0375 m s^-2. The time taken to decelerate to rest is given by:

\(0 = 15 - 0.0375 t_3\)

\(t_3 = \frac{15}{0.0375} = 400 \text{ s}\)

The total time is:

\(600 + 2600 + 400 = 3600 \text{ s}\)

(ii) The velocity-time graph consists of three segments: an upward slope for 600 s, a horizontal line for 2600 s, and a downward slope for 400 s. The distance is the area under the graph:

\(d = \frac{1}{2} \times 600 \times 15 + 2600 \times 15 + \frac{1}{2} \times 400 \times 15\)

\(d = 4500 + 39000 + 3000 = 46500 \text{ m}\)

(iii) The speed of the train is 7.5 m s^-1 at two points: during acceleration and deceleration. During acceleration:

\(7.5 = 0.025 t\)

\(t = \frac{7.5}{0.025} = 300 \text{ s}\)

During deceleration:

\(7.5 = 15 - 0.0375 (t - 3200)\)

\(0.0375 (t - 3200) = 7.5\)

\(t - 3200 = \frac{7.5}{0.0375} = 200\)

\(t = 3400 \text{ s}\)

(i) For 0 < t < 4, the acceleration is 0.75 m s^-2. The initial velocity is 0, so the velocity at t = 4 is given by:

\(v(4) = 0 + 0.75 \times 4 = 3 \text{ m s}^{-1}\)

For 4 < t < 54, the acceleration is 0, so the velocity remains constant at 3 m s^-1.

\(For 54 < t < 60, the acceleration is -0.5 m s^-2. The velocity at t = 60 is given by:\)

\(v(60) = v(54) + (-0.5) \times (60 - 54) = 3 + (-0.5) \times 6 = 0 \text{ m s}^{-1}\)

The velocity-time graph consists of three segments: a line with positive slope from (0,0) to (4,3), a horizontal line from (4,3) to (54,3), and a line with negative slope from (54,3) to (60,0).

(ii) The distance XY can be found by calculating the area under the velocity-time graph:

Area of the first segment (triangle):

\(\frac{1}{2} \times 4 \times 3 = 6 \text{ m}\)

Area of the second segment (rectangle):

\(50 \times 3 = 150 \text{ m}\)

Area of the third segment (triangle):

\(\frac{1}{2} \times 6 \times 3 = 9 \text{ m}\)

Total distance XY is:

\(6 + 150 - 9 = 165 \text{ m}\)