(i) To find the maximum velocity, set the acceleration to zero:

\(25 - t^2 = 0\)

\(t^2 = 25\)

\(t = 5\)

Integrate the acceleration to find the velocity:

\(v = \int (25 - t^2) \, dt = 25t - \frac{1}{3}t^3 + C\)

Since the particle is initially at rest, \(v = 0\) when \(t = 0\), so \(C = 0\).

Substitute \(t = 5\) to find the maximum velocity:

\(v = 25 \times 5 - \frac{1}{3} \times 5^3 = 125 - \frac{125}{3} = \frac{250}{3} = 83\frac{1}{3} \text{ m s}^{-1}\)

(ii) To find the total distance travelled, integrate the velocity function from \(t = 0\) to \(t = 5\):

\(s = \int_0^5 (25t - \frac{1}{3}t^3) \, dt = \left[ \frac{25}{2}t^2 - \frac{1}{12}t^4 \right]_0^5\)

\(s = \left( \frac{25}{2} \times 25 - \frac{1}{12} \times 625 \right) = 260 \text{ m}\)

(iii) For \(t > 9\), the acceleration is given by \(a = -3t^{-1/2}\). Integrate to find the velocity change from \(t = 9\) to \(t = 25\):

At \(t = 9\), \(v = 25 \times 9 - \frac{1}{3} \times 9^3 = -18 \text{ m s}^{-1}\).

Integrate the acceleration:

\(\Delta v = \int_9^{25} -3t^{-1/2} \, dt = \left[ -6t^{1/2} \right]_9^{25}\)

\(\Delta v = -6 \times 5 + 6 \times 3 = -30 \text{ m s}^{-1}\)

Thus, the velocity at \(t = 25\) is \(-18 - 12 = -30 \text{ m s}^{-1}\).