(a) To find the value of k, integrate the acceleration to find the velocity:

\(v = \int (2t - \frac{3}{5}t^2) \, dt = t^2 - \frac{1}{5}t^3 + C\)

Since the cyclist starts from rest, \(v = 0\) when \(t = 0\), so \(C = 0\).

Set \(v = 0\) to find \(k\):

\(t^2 - \frac{1}{5}t^3 = 0\)

\(t^2(1 - \frac{1}{5}t) = 0\)

\(t = 0\) or \(t = 5\)

Since \(t = 0\) is the starting point, \(k = 10\).

(b) The maximum speed occurs when acceleration \(a = 0\):

\(2t - \frac{3}{5}t^2 = 0\)

\(t(2 - \frac{3}{5}t) = 0\)

\(t = 0\) or \(t = \frac{10}{3}\)

Substitute \(t = \frac{10}{3}\) into the velocity equation:

\(v = (\frac{10}{3})^2 - \frac{1}{5}(\frac{10}{3})^3\)

\(v = \frac{100}{9} - \frac{1000}{135}\)

\(v = \frac{800}{135} = \frac{8\sqrt{30}}{9} \approx 4.87 \text{ m/s}\)

(c) Integrate the velocity to find displacement:

\(s = \int (t^2 - \frac{1}{5}t^3) \, dt = \frac{t^3}{3} - \frac{t^4}{20} + C\)

Substitute \(t = \frac{10}{3}\) and \(t = 10\) to find the distance:

\(s(\frac{10}{3}) = \frac{(\frac{10}{3})^3}{3} - \frac{(\frac{10}{3})^4}{20}\)

\(s(10) = \frac{10^3}{3} - \frac{10^4}{20}\)

Distance = \(s(10) - s(\frac{10}{3}) \approx 20.7 \text{ m}\)