(i) The displacement of Q is given by integrating the velocity function:

\(s_Q = \int v_Q \, dt = \int (3t - 0.3t^2) \, dt = 1.5t^2 - 0.1t^3 + C\).

Using the limits from 0 to 10, we find:

\(s_Q(10) = 1.5(10)^2 - 0.1(10)^3 = 150 - 100 = 50\).

The displacement of P is the area under the velocity-time graph, which is a triangle with base 10 and height 10:

\(s_P = \frac{1}{2} \times 10 \times 10 = 50\).

Since \(s_P = s_Q\), the greatest velocity of P is 10 m s^-1.

(ii) The acceleration of P is constant during the interval \(0 < t < 5\) and is given by:

\(a_P = \frac{10}{5} = 2 \text{ m s}^{-2}\).

The acceleration of Q is found by differentiating the velocity function:

\(a_Q = \frac{d}{dt}(3t - 0.3t^2) = 3 - 0.6t\).

Setting \(a_Q = a_P\):

\(3 - 0.6t = 2\).

Solving for \(t\):

\(0.6t = 1\).

\(t = \frac{1}{0.6} = 1.67\) (or \(1\frac{2}{3}\)).