(a) To find the displacement, integrate the velocity function:

\(s = \int (t^2 - 8t^{3/2} + 10t) \, dt\)

\(s = \left[ \frac{t^3}{3} - \frac{16}{5}t^{5/2} + 5t^2 \right] + C\)

Using the limits from \(t = 0\) to \(t = 1\),

\(s = \left( \frac{1^3}{3} - \frac{16}{5} \times 1^{5/2} + 5 \times 1^2 \right) - \left( \frac{0^3}{3} - \frac{16}{5} \times 0^{5/2} + 5 \times 0^2 \right)\)

\(s = \frac{1}{3} - \frac{16}{5} + 5 = \frac{32}{15} \approx 2.13 \text{ m}\)

(b) To find the minimum velocity, differentiate the velocity function:

\(a = \frac{dv}{dt} = 2t - 12t^{1/2} + 10\)

Set \(a = 0\):

\(2t - 12t^{1/2} + 10 = 0\)

Solving this quadratic equation in terms of \(\sqrt{t}\),

\(2(\sqrt{t} - 5)(\sqrt{t} - 1) = 0\)

\(t = 1\) or \(t = 25\)

Check the nature of the stationary point by differentiating \(a\):

\(\frac{da}{dt} = 2 - 6t^{-1/2}\)

At \(t = 25\), \(\frac{da}{dt} = 0.8\), indicating a minimum.

Evaluate \(v\) at \(t = 25\):

\(v = 25^2 - 8 \times 25^{3/2} + 10 \times 25 = -125 \text{ m s}^{-1}\)