(i) For 0 ≤ t ≤ 5, the velocity is given by v = t - 0.1t². The distance travelled, s_1, is the integral of velocity with respect to time:

\(s_1 = \int_0^5 (t - 0.1t^2) \, dt = \left[ \frac{t^2}{2} - \frac{0.1t^3}{3} \right]_0^5\)

\(s_1 = \left( \frac{25}{2} - \frac{0.1 \times 125}{3} \right) = 8.33 \text{ m}\)

(ii) For 5 ≤ t ≤ 45, the velocity is constant. Let the constant velocity be v_c. Since the particle starts from rest and returns to rest, the symmetry implies v_c = 2.5 m/s. The distance travelled, s_2, is:

\(s_2 = v_c \times (45 - 5) = 2.5 \times 40 = 100 \text{ m}\)

\(For 45 ≤ t ≤ 50, the velocity is given by v = 9t - 0.1t^2 - 200. The distance travelled, s_3, is:\)

\(s_3 = \int_{45}^{50} (9t - 0.1t^2 - 200) \, dt\)

\(s_3 = \left[ \frac{9t^2}{2} - \frac{0.1t^3}{3} - 200t \right]_{45}^{50}\)

\(s_3 = \left( \frac{9 \times 2500}{2} - \frac{0.1 \times 125000}{3} - 200 \times 50 \right) - \left( \frac{9 \times 2025}{2} - \frac{0.1 \times 91125}{3} - 200 \times 45 \right)\)

\(s_3 = 8.33 \text{ m}\)

The total distance from O to A is:

\(s = s_1 + s_2 + s_3 = 8.33 + 100 + 8.33 = 117 \text{ m}\)

The average speed is:

\(\text{Average speed} = \frac{\text{Total distance}}{\text{Total time}} = \frac{117}{50} = 2.33 \text{ m/s}\)