(i) Using the equation of motion:

\(v = u + at\)

where \(v = 2.2\) m s^-1, \(u = 1.8\) m s^-1, and \(a = 0.004\) m s^-2.

Substitute the values:

\(2.2 = 1.8 + 0.004t\)

Solve for \(t\):

\(0.004t = 0.4\)

\(t = 100 \text{ s}\)

To find the distance \(s\), use:

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

Substitute the values:

\(s = 1.8 \times 100 + \frac{1}{2} \times 0.004 \times 100^2\)

\(s = 180 + 20 = 200 \text{ m}\)

(ii) (a) Integrate the velocity function to find the displacement \(s(t)\):

\(s = \int k(200t - t^2) \, dt = k \left( 100t^2 - \frac{t^3}{3} \right) + C\)

Given \(s(0) = 0\), \(C = 0\).

At \(t = 100\), \(s(100) = 200\):

\(k \left( 100 \times 100^2 - \frac{100^3}{3} \right) = 200\)

Solve for \(k\):

\(k \times 666666.67 = 200\)

\(k = 0.0003\)

(b) The cyclist’s speed at \(t = 100\) is:

\(v = k(200 \times 100 - 100^2)\)

\(v = 0.0003 \times 10000 = 3 \text{ m s}^{-1}\)

(iii) The velocity-time graph for the man is a straight line with positive slope, starting at 1.8 m s^-1 and ending at 2.2 m s^-1. The cyclist’s graph is a parabola with decreasing slope, starting at 0 m s^-1 and ending at 3 m s^-1.