(i) (a) The acceleration of P from A to B is given by \(g \sin \alpha\). Substituting \(g = 9.8\) m s^-2 and \(\sin \alpha = 0.28\), we have:

\(a = 9.8 \times 0.28 = 2.8 \text{ m s}^{-2}\)

(b) For the motion from B to C, using Newton's second law:

\(mg \times 0.28 - 0.5mg \times 0.96 = ma\)

Solving for \(a\):

\(a = \frac{mg(0.28 - 0.5 \times 0.96)}{m} = -2 \text{ m s}^{-2}\)

(ii) Using \(v^2 = u^2 + 2as\) for AB and BC and \(AB + BC = 5\):

\(v_B^2 = 2 \times 2.8(AB)\)

\(2^2 = 5.6(AB) - 2 \times 2(5 - AB)\)

Solving gives \(AB = 2.5 \text{ m}\).

(iii) Using \(t = \frac{2s}{u+v}\) for AB and BC:

\(T = 2 \times 2.5 \div (0 + \sqrt{14}) + 2 \times 2.5 \div (\sqrt{14} + 2)\)

The time taken is 2.21 s.