(i) The acceleration \(a\) is given by \(a = \frac{dv}{dt} = 0.5 - 0.02t\). Setting \(a = 0.1\), we solve \(0.5 - 0.02t = 0.1\) to find \(t = 20\) s.

(ii) The initial velocity at B is \(u = 0.5 \times 20 - 0.01 \times 20^2 = 6\) m s\(^{-1}\). Using \(v = u + at\), where \(v = 14\) m s\(^{-1}\) and \(a = 0.1\) m s\(^{-2}\), we have \(14 = 6 + 0.1t\). Solving gives \(t = 80\) s.

(iii) Using \(v^2 = u^2 + 2as\) for the section CD, where \(u = 14\) m s\(^{-1}\), \(a = -0.3\) m s\(^{-2}\), and \(s = 300\) m, we find \(v^2 = 14^2 - 2 \times 0.3 \times 300\). Solving gives \(v = 4\) m s\(^{-1}\).

(iv) The displacement \(s\) is given by integrating the velocity function: \(s = \int v \, dt = 0.25t^2 - 0.01t^3/3 + C\). For AB, \(s = 0.25 \times 20^2 - 0.01 \times 20^3/3 = 73.3\) m. For BC, \(s = \frac{1}{2}(6 + 14) \times 80 = 800\) m. Thus, the total distance AD is \(73.3 + 800 + 300 = 1170\) m.