(a) To find the displacement \(s\), integrate the velocity function:

\(s = \int k(t^2 - 10t + 21) \, dt = k \left( \frac{1}{3} t^3 + 5t^2 + 21t \right) + C\)

Using the given displacements:

\(2.85 = k \left( \frac{1}{3} \times 3^3 - 5 \times 3^2 + 21 \times 3 \right) + C\)

\(2.4 = k \left( \frac{1}{3} \times 6^3 - 5 \times 6^2 + 21 \times 6 \right) + C\)

Solving these equations gives \(k = 0.05\) and \(C = 1.5\).

Thus, the displacement is:

\(s = 0.05 \left( \frac{1}{3} t^3 - 5t^2 + 21t \right) + 1.5\)

(b) To find when the velocity is a minimum, differentiate \(v\):

\(a = \frac{dv}{dt} = 0.05(2t - 10)\)

Set \(a = 0\) to find \(t = 5\).

Substitute \(t = 5\) into the displacement equation:

\(s = 0.05 \left( \frac{1}{3} \times 5^3 - 5 \times 5^2 + 21 \times 5 \right) + 1.5 = 2.58 \text{ m}\)