To find when the particle is at rest, we need to find when the velocity \(v = \frac{ds}{dt}\) is zero.

Differentiate \(s = t^{\frac{5}{2}} - \frac{15}{4} t^{\frac{3}{2}} + 6\) with respect to \(t\):

\(v = \frac{5}{2} t^{\frac{3}{2}} - \frac{45}{8} t^{\frac{1}{2}}\).

Set \(v = 0\):

\(\frac{5}{2} t^{\frac{3}{2}} - \frac{45}{8} t^{\frac{1}{2}} = 0\).

Factor out \(t^{\frac{1}{2}}\):

\(t^{\frac{1}{2}} \left( \frac{5}{2} t - \frac{45}{8} \right) = 0\).

Since \(t^{\frac{1}{2}} = 0\) is not possible (as \(t = 0\) is the starting point), solve:

\(\frac{5}{2} t - \frac{45}{8} = 0\).

\(5t = \frac{45}{4}\).

\(t = \frac{45}{20} = 2.25\).

Substitute \(t = 2.25\) back into the displacement equation:

\(s = (2.25)^{\frac{5}{2}} - \frac{15}{4} (2.25)^{\frac{3}{2}} + 6\).

Calculate \(s\):

\(s = \frac{15}{16}\).