(i) For the first 8 seconds, the velocity \(v(8) = 0.25 \times 8 = 2\) m/s. Using the given velocity equation:

\(v = -0.1t^2 + 2.4t - k\)

Substitute \(t = 8\):

\(2 = -0.1(8)^2 + 2.4(8) - k\)

\(2 = -6.4 + 19.2 - k\)

\(k = 10.8\)

(ii) To find the maximum velocity, differentiate \(v\) with respect to \(t\) and set the derivative to zero:

\(\frac{dv}{dt} = -0.2t + 2.4\)

Set \(\frac{dv}{dt} = 0\):

\(-0.2t + 2.4 = 0\)

\(t = 12\)

Substitute \(t = 12\) into the velocity equation:

\(v = -0.1(12)^2 + 2.4(12) - 10.8\)

\(v = -14.4 + 28.8 - 10.8\)

\(v = 3.6\) m/s

(iii) Displacement for the first 8 seconds:

\(s_1 = \frac{1}{2} \times 0.25 \times 8^2 = 8\) m

Displacement from \(t = 8\) to \(t = 18\):

\(s_2 = \int_{8}^{18} (-0.1t^2 + 2.4t - 10.8) \, dt\)

\(s_2 = \left[ -0.1 \frac{t^3}{3} + 1.2t^2 - 10.8t \right]_{8}^{18}\)

\(s_2 = [-0.1(18)^3/3 + 1.2(18)^2 - 10.8(18)] - [-0.1(8)^3/3 + 1.2(8)^2 - 10.8(8)]\)

\(s_2 = 26.7\) m

Total displacement \(s = s_1 + s_2 = 8 + 26.7 = 34.7\) m