(i) For cyclist P, using the equation of motion:

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

where \(u = 3\) m s^-1, \(a = 0.005\) m s^-2, and \(t = 400\) s:

\(AB = 3 \times 400 + \frac{1}{2} \times 0.005 \times 400^2 = 1600 \text{ m}\)

To find P's speed at B:

\(v_B = u + at = 3 + 0.005 \times 400 = 5 \text{ m s}^{-1}\)

(ii) For cyclist Q, integrate the velocity function to find the distance:

\(\int_0^{400} v \, dt = 1600\)

\(\int_0^{400} (0.04t - 0.0001t^2 + k) \, dt = 1600\)

\([0.02t^2 - 0.0001 \frac{t^3}{3} + kt]_0^{400} = 1600\)

\(400k = 1600 - 0.02 \times 400^2 + 0.0001 \times \frac{400^3}{3}\)

\(k = 4 - \frac{8 + 16/3}{3} = \frac{4}{3}\)

To find the maximum speed, differentiate \(v\) and set \(\frac{dv}{dt} = 0\):

\(\frac{dv}{dt} = 0.04 - 0.0002t = 0\)

\(t = 200\) s, \(v_{\text{max}} = 0.04 \times 200 - 0.0001 \times 200^2 + \frac{4}{3} = 5.33 \text{ m s}^{-1}\)

(iii) For cyclist P, time taken from B to C:

\(\text{Time} = \frac{1400}{5} = 280 \text{ s}\)

For cyclist Q, using \(s = ut + \frac{1}{2}at^2\):

\(1400 = \frac{4}{3} \times 280 + \frac{1}{2} \times a \times 280^2\)

\(a = 0.0262 \text{ m s}^{-2}\)