(a) To find when the particle is at rest, set \(v = 0\):

\(2t^2 - 5t + 3 = 0\).

Factorize to get \((2t - 3)(t - 1) = 0\), giving \(t = 1\) or \(t = 1.5\).

To find the minimum velocity, complete the square or use calculus. The minimum occurs at \(t = 1.25\):

\(v = 2\left(\frac{5}{4}\right)^2 - 5\left(\frac{5}{4}\right) + 3 = -0.125\) m s\(^{-1}\).

(b) The velocity-time graph is a quadratic curve with roots at \(t = 1\) and \(t = 1.5\). Key points are (1.25, -0.125), (0, 3), (1, 0), (1.5, 0), (3, 6).

(c) To find the distance, integrate the velocity function from \(t = 1\) to \(t = 1.5\):

\(s = \int_{1}^{1.5} (2t^2 - 5t + 3) \, dt\).

Calculate:

\(s = \left[ \frac{2}{3}t^3 - \frac{5}{2}t^2 + 3t \right]_{1}^{1.5}\).

\(s = \left[ \frac{2}{3}(1.5)^3 - \frac{5}{2}(1.5)^2 + 3(1.5) \right] - \left[ \frac{2}{3}(1)^3 - \frac{5}{2}(1)^2 + 3(1) \right]\).

\(s = 0.0417\) m.