(i) The speed of P at A is given by the formula for final velocity:

\(v = u + at\)

where \(u = 1.3\) m s^-1, \(a = 0.1\) m s^-2, and \(t = 20\) s.

\(v = 1.3 + 0.1 \times 20 = 3.3\) m s^-1

The speed of Q at B is the same as the speed of P at A, so \(v_Q(20) = 3.3\) m s^-1.

The acceleration of Q is \(0.016t\), so the velocity function is:

\(v_Q(t) = \int 0.016t \, dt = 0.008t^2 + v_Q(0)\)

Substituting \(t = 20\) and \(v_Q(20) = 3.3\):

\(3.3 = 0.008 \times 20^2 + v_Q(0)\)

\(3.3 = 3.2 + v_Q(0)\)

\(v_Q(0) = 0.1\) m s^-1

(ii) The distance AO is calculated using the formula for distance under constant acceleration:

\(s = ut + \frac{1}{2}at^2\)

\(AO = 1.3 \times 20 + \frac{1}{2} \times 0.1 \times 20^2 = 46\) m

The distance OB is calculated using the integral of velocity:

\(OB = \int (0.008t^2 + 0.1) \, dt\) from 0 to 20

\(OB = \left[ \frac{0.008}{3}t^3 + 0.1t \right]_0^{20}\)

\(OB = \frac{0.008}{3} \times 20^3 + 0.1 \times 20 = 23.3\) m

Therefore, the total distance \(AB = AO + OB = 46 + 23.3 = 69.3\) m