(i) For \(t = 3\), substitute into the velocity equation: \(v = -0.2(3)^2 + 4(3) - 15 = -4.8\) m s^-1. The acceleration for the first 3 s is constant, using \(v = u + at\) with \(u = 0\), \(v = -4.8\), and \(t = 3\), gives \(a = \frac{-4.8}{3} = -1.6\) m s^-2.

(ii) To find the maximum velocity, set \(\frac{dv}{dt} = 0\). Differentiate \(v = -0.2t^2 + 4t - 15\) to get \(\frac{dv}{dt} = -0.4t + 4\). Solve \(-0.4t + 4 = 0\) to find \(t = 10\). Substitute \(t = 10\) into the velocity equation: \(v = -0.2(10)^2 + 4(10) - 15 = 5\) m s^-1.

(iii) (a) Distance from 0 to 3 s is \(\frac{1}{2} \times 3 \times 4.8 = 7.2\) m. From 3 to 5 s, integrate \(v = -0.2t^2 + 4t - 15\) from 3 to 5: \(\int_{3}^{5} (-0.2t^2 + 4t - 15) \, dt = 4.533\) m. Total distance = 7.2 + 4.533 = 11.733 m. Average speed = \(\frac{11.733}{5} = 2.35\) m s^-1.

(b) For the whole journey, integrate from 3 to 15: \(\int_{3}^{15} (-0.2t^2 + 4t - 15) \, dt = 45.066\) m. Total distance = 7.2 + 45.066 = 52.266 m. Average speed = \(\frac{52.266}{15} = 3.00\) m s^-1.