(a) To find the velocity at \(t = 4\), integrate the acceleration \(a = 0.3t^{\frac{1}{2}}\) from \(t = 0\) to \(t = 4\):

\(v = \int 0.3t^{\frac{1}{2}} \, dt = 0.3 \cdot \frac{2}{3} t^{\frac{3}{2}} + c = 0.2t^{\frac{3}{2}} + c\).

Since the particle starts from rest, \(c = 0\). Thus, \(v = 0.2 \times 4^{\frac{3}{2}} = 1.6 \text{ m/s}\).

(b) For \(4 < t \leq T\), \(a = -kt^{-\frac{3}{2}}\). Integrate to find velocity:

\(v = \int -kt^{-\frac{3}{2}} \, dt = -k \cdot \frac{-2}{1} t^{-\frac{1}{2}} + d = 2kt^{-\frac{1}{2}} + d\).

Given \(v = 0.3 \text{ m/s}\) at \(t = 16\), and no change at \(t = 4\), equate velocities:

\(1.6 = 2k \cdot 4^{-\frac{1}{2}} + d\) and \(0.3 = 2k \cdot 16^{-\frac{1}{2}} + d\).

Solving gives \(k = 2.6\) and \(v = 5.2t^{-\frac{1}{2}} - 1\).

(c) Set \(v = 0\) for \(t = T\):

\(5.2T^{-\frac{1}{2}} - 1 = 0\).

Solving gives \(T = \frac{676}{25}\) or 27.04.

(d) Total distance is the integral of speed from \(t = 0\) to \(t = T\):

\(\int_0^4 0.2t^{\frac{1}{2}} \, dt + \int_4^{27.04} (5.2t^{-\frac{1}{2}} - 1) \, dt\).

Calculating gives 12.8 m.