(i) To find the velocity immediately before the particle hits the wall at \(t = 60\), substitute \(t = 60\) into \(v = 0.05t - 0.0005t^2\):

\(v = 0.05(60) - 0.0005(60)^2 = 3 - 1.8 = 1.2 \text{ ms}^{-1}\).

Immediately after hitting the wall, the velocity is given as \(-1 \text{ ms}^{-1}\).

(ii) The total distance travelled is the sum of the distances from \(O\) to the wall \(W\) and from \(W\) to \(A\).

Distance \(OW = \int_{0}^{60} (0.05t - 0.0005t^2) \, dt = \left[ 0.025t^2 - 0.0005 \frac{t^3}{3} \right]_{0}^{60} = 54 \text{ m}\).

Distance \(WA = \int_{60}^{100} (0.025t - 2.5) \, dt = \left[ 0.0125t^2 - 2.5t \right]_{60}^{100} = 20 \text{ m}\).

Total distance = \(54 + 20 = 74 \text{ m}\).

(iii) To find the maximum speed, consider the velocity function \(v = 0.05t - 0.0005t^2\) for \(0 \leq t < 60\). Differentiate to find the critical points:

\(\frac{dv}{dt} = 0.05 - 0.001t = 0 \Rightarrow t = 50\).

Substitute \(t = 50\) into the velocity equation:

\(v = 0.05(50) - 0.0005(50)^2 = 1.25 \text{ ms}^{-1}\).

The maximum speed is \(1.25 \text{ ms}^{-1}\) at \(t = 50\).

The velocity-time graph is a quadratic curve from \((0,0)\) to \((60,1.2)\) with a maximum at \((50,1.25)\), followed by a straight line from \((60,-1)\) to \((100,0)\).