(i) For the cyclist's initial acceleration phase, use the equation:

\(s = ut + \frac{1}{2}at^2\)

\(36 = 0 + \frac{1}{2} \times 0.5 \times t^2\)

\(t^2 = 144\)

\(t = 12 \text{ s}\)

The speed at the end of this phase is found using:

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

\(v^2 = 0 + 2 \times 0.5 \times 36\)

\(v = 6 \text{ m/s}\)

The cyclist then travels at constant speed for 25 s, covering:

\(s = vt = 6 \times 25 = 150 \text{ m}\)

The remaining distance to B is:

\(210 - 36 - 150 = 24 \text{ m}\)

Using the equation for deceleration:

\(s = \frac{(u+v)}{2}t\)

\(24 = \frac{(6+0)}{2}t\)

\(t = 8 \text{ s}\)

\(Total time = 12 + 25 + 8 = 45 s\)

(ii) Distance travelled by cyclist after 12 s is:

\(36 + 6(t - 12)\)

Distance travelled by car is:

\(0.5 \times 4 \times (t - 24)^2\)

Equating the distances:

\(2(t - 24)^2 = 36 + 6(t - 12)\)

\(2t^2 - 96t + 1152 = 36 + 6t - 72\)

\(2t^2 - 96t + 1152 = 6t - 36\)

\(2t^2 - 102t + 1188 = 0\)

Solving the quadratic equation:

\(t = 33 \text{ or } t = 18\)

The correct solution is \(t = 33 \text{ s}\)