(i) For \(0 < t < 0.5\), the average acceleration \(a_{\text{ave}}\) is calculated as:

\(a_{\text{ave}} = \frac{6 - 4}{0.5 - 0} = \frac{2}{0.5} = 4 \text{ m/s}^2\)

For \(16.3 < t < 24.5\), the average acceleration \(a_{\text{ave}}\) is:

\(a_{\text{ave}} = \frac{21 - 19}{24.5 - 16.3} = \frac{2}{8.2} \approx 0.244 \text{ m/s}^2\)

(ii) Using the accelerations from part (i), assume \(a_1 = 4 \text{ m/s}^2\) at \(v = 5\) and \(a_2 = 0.244 \text{ m/s}^2\) at \(v = 20\).

Using Newton's second law, \(F = ma\), and the given power and resistance:

For \(v = 5\):

\(\frac{P}{5} - R = 1230 \times 4\)

For \(v = 20\):

\(\frac{P}{20} - R = 1230 \times 0.244\)

Solving these equations simultaneously:

\(P = 30800 \text{ W}\)

\(R = 1240 \text{ N}\)