(i) For the section AB, using the equation of motion:

\(s_{AB} = ut + \frac{1}{2} a t^2\)

where \(u = 14\) m s^-1, \(t = 5\) s, and \(a\) is the acceleration.

\(s_{AB} = 14 \times 5 + \frac{1}{2} a \times 5^2\)

\(s_{AB} = 70 + \frac{25}{2} a\)

For the section AC, the time is \(5 + 3 = 8\) s:

\(s_{AC} = 14 \times 8 + \frac{1}{2} a \times 8^2\)

\(s_{AC} = 112 + 32a\)

Since \(AC = 2AB\), we have:

\(112 + 32a = 2(70 + 12.5a)\)

\(112 + 32a = 140 + 25a\)

\(7a = 28\)

\(a = 4 \text{ m s}^{-2}\)

(ii) To find the speed at point C, use the equation:

\(v = u + at\)

where \(u = 14\) m s^-1, \(a = 4\) m s^-2, and \(t = 8\) s:

\(v = 14 + 4 \times 8\)

\(v = 46 \text{ m s}^{-1}\)