(i) Using Newton's second law, the frictional force is given by:

\(F = \mu R = \mu mg\)

where \(\mu\) is the coefficient of friction, \(R\) is the normal reaction, and \(m\) is the mass of the particle. The deceleration \(a\) is given by:

\(\mu mg = ma\)

Thus, \(a = \mu g\).

For particle P, \(\mu = 0.2\):

\(a_P = 0.2 \times 9.8 = 2 \text{ m s}^{-2}\)

For particle Q, \(\mu = 0.25\):

\(a_Q = 0.25 \times 9.8 = 2.5 \text{ m s}^{-2}\)

(ii) Using the equation of motion \(s = ut + \frac{1}{2}at^2\), where \(s_P = s_Q + 5\):

For P:

\(s_P = 8t - \frac{1}{2} \times 2 \times t^2 = 8t - t^2\)

For Q:

\(s_Q = 3t - \frac{1}{2} \times 2.5 \times t^2 = 3t - 1.25t^2\)

Equating and solving:

\(8t - t^2 = 3t - 1.25t^2 + 5\)

\(5t = 0.25t^2 + 5\)

\(t^2 - 5t + 5 = 0\)

Solving for \(t\):

\(t = \sqrt{120} - 10 \approx 0.95445\)

Using \(v = u + at\) for both P and Q:

For P:

\(v_P = 8 - 2 \times 0.95445 = 6.09 \text{ m s}^{-1}\)

For Q:

\(v_Q = 3 - 2.5 \times 0.95445 = 0.614 \text{ m s}^{-1}\)