(a) To find when the particle is at rest, set the velocity \(v = \frac{ds}{dt}\) to zero. For \(0 \leq t \leq 6\), \(s = t^2 - 3t + 2\).

Differentiate: \(v = \frac{ds}{dt} = 2t - 3\).

Set \(v = 0\): \(2t - 3 = 0 \Rightarrow t = 1.5\).

(b) At \(t = 6\), find the velocity using \(s = \frac{24}{t} - \frac{t^2}{4} + 25\) for \(t \geq 6\).

Differentiate: \(v = -\frac{24}{t^2} - 0.5t\).

At \(t = 6\), \(v = -\frac{24}{6^2} - 0.5 \times 6 = 9 \text{ m/s}\).

Velocity when leaves: \(v = -3.67 \text{ m/s}\).

(c) Calculate displacement at key points:

• At \(t = 0\), \(s = 2\).
• At \(t = 1.5\), \(s = -0.25\).
• At \(t = 6\), \(s = 20\).
• At \(t = 10\), \(s = 2.4\).

Total distance = \(2 + 0.25 + 0.25 + 20 + (20 - 2.4) = 40.1 \text{ m}\).