(i) Let the speed of Q at time t be v_Q. Then, using the equation of motion,

\(v_Q = 3 + 2t\).

The speed of P is 1.8 times the speed of Q, so

\(v_P = 1.8v_Q = 1.8(3 + 2t) = 5 + 4t\).

Equating the expressions for \(v_P\), we have:

\(5 + 4t = 1.8(3 + 2t)\).

Solving for \(t\):

\(5 + 4t = 5.4 + 3.6t\)

\(0.4t = 0.4\)

\(t = 1\).

Substitute \(t = 1\) into \(v_P = 5 + 4t\):

\(v_P = 5 + 4(1) = 9 \text{ m s}^{-1}\).

(ii) Using the equation \(s = ut + \frac{1}{2}at^2\) for both particles:

For P: \(s_P = 5t + \frac{1}{2} \times 4 \times t^2\)

For Q: \(s_Q = 3t + \frac{1}{2} \times 2 \times t^2\)

Since \(s_P + s_Q = 51\):

\(5t + 2t^2 + 3t + t^2 = 51\)

\(3t^2 + 8t - 51 = 0\).

Solving the quadratic equation:

\((3t + 17)(t - 3) = 0\)

\(t = 3 \text{ s}\) (since time cannot be negative).