(i) Using the equation of motion:

\(v = u + at\)

where \(v = 1.3\) m/s, \(u = 0.9\) m/s, and \(a = 0.004\) m/s², we have:

\(1.3 = 0.9 + 0.004T\)

Solving for \(T\):

\(T = \frac{1.3 - 0.9}{0.004} = 100 \text{ s}\)

Using the equation:

\(v^2 = u^2 + 2as\)

where \(s\) is the distance, we have:

\(1.3^2 = 0.9^2 + 2 \times 0.004 \times s\)

Solving for \(s\):

\(s = \frac{1.3^2 - 0.9^2}{2 \times 0.004} = 110 \text{ m}\)

(ii) The cyclist's speed is given by:

\(v_c = kt^3\)

Integrating to find the distance:

\(\int kt^3 \, dt = \frac{1}{4} kt^4\)

Setting the limits from 0 to 100:

\(\frac{1}{4} k (100)^4 = 110\)

Solving for \(k\):

\(k = \frac{110 \times 4}{100^4} = 4.4 \times 10^{-6}\)

At \(t = 64.05\):

\(v_w = 0.9 + 0.004 \times 64.05\)

\(v_c = 4.4 \times 10^{-6} \times (64.05)^3\)

Both speeds are:

\(v_w = v_c = 1.16 \text{ m/s}\)

(iii) The cyclist's acceleration is:

\(a_c = \frac{d}{dt}(kt^3) = 3kt^2\)

At \(t = 100\):

\(a_c = 3 \times 4.4 \times 10^{-6} \times 100^2 = 0.132 \text{ m/s}^2\)