(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:

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