(a) We have the equations from the given conditions:

\(4b + 4^{\frac{3}{2}}c = 8\)

\(9b + 9^{\frac{3}{2}}c = 13.5\)

Solving these simultaneously, we find \(b = 3\) and \(c = -0.5\).

(b) The acceleration \(a\) is given by \(a = \frac{dv}{dt} = b + \frac{3}{2}ct^{\frac{1}{2}}\).

Substituting \(b = 3\), \(c = -0.5\), and \(t = 1\), we get:

\(a = 3 + \frac{3}{2}(-0.5)(1)^{\frac{1}{2}} = 3 - 0.75 = 2.25 \text{ m/s}^2\)

(c) For instantaneous rest, set \(v = 0\):

\(3t - 0.5t^{\frac{3}{2}} = 0\)

Solving gives \(t = 36\).

To find the distance, integrate \(v\) to get \(s\):

\(s = \int (3t - 0.5t^{\frac{3}{2}}) \, dt = \frac{3}{2}t^2 - \frac{1}{5}t^{\frac{5}{2}} + C\)

Using limits from 0 to 36, \(s = 388.8 \text{ m}\).

(d) For the speed when returning to \(O\), solve:

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

\(t = 56.25\).

Substitute back into \(v\):

\(v = 3(56.25) - 0.5(56.25)^{\frac{3}{2}} = 42.2 \text{ m/s}\)