(a) At \(t = 5\), the velocity \(v = 0\):

\(0 = \frac{9}{4} + \frac{b}{(5+1)^2} - c \times 5^2\)

\(0 = \frac{9}{4} + \frac{b}{36} - 25c\)

For acceleration \(a = \frac{dv}{dt}\):

\(a = -\frac{2b}{(t+1)^3} - 2ct\)

At \(t = 5\), \(a = -\frac{13}{12}\):

\(-\frac{13}{12} = -\frac{2b}{(5+1)^3} - 2c \times 5\)

\(-\frac{13}{12} = -\frac{b}{108} - 10c\)

Solving the simultaneous equations:

\(\frac{b}{36} - 25c = -\frac{9}{4}\)

\(-\frac{b}{108} - 10c = -\frac{13}{12}\)

Leads to \(b = 9\) and \(c = 0.1\).

(b) To find the distance travelled, integrate the velocity:

\(\int \left( \frac{9}{4} + \frac{9}{(t+1)^2} - 0.1t^2 \right) \, dt\)

\(= \left[ \frac{9}{4}t - \frac{9}{t+1} - \frac{1}{30}t^3 \right]_0^{10}\)

\(= \left[ \frac{9}{4} \times 10 - \frac{9}{11} - \frac{1}{30} \times 10^3 \right] - \left[ \frac{9}{4} \times 5 - \frac{9}{6} - \frac{1}{30} \times 5^3 \right]\)

\(= 5.583 + 9 + 11.651 + 5.583\)

\(= 31.8 \text{ m}\)