(a) The acceleration is given by \(a = 16k - kt^2\). The velocity \(v\) is the integral of acceleration: \(v = \int a \, dt = 16kt - \frac{1}{3} kt^3\).

Given \(v = 8\) when \(t = 4\), substitute to find \(k\):

\(8 = 16k \times 4 - \frac{1}{3} k \times 4^3\)

Solve for \(k\): \(k = \frac{3}{16}\).

Now, find \(s\) by integrating \(v\):

\(s = \int v \, dt = 8kt^2 - \frac{1}{12} kt^4\)

Substitute \(k = \frac{3}{16}\):

\(s = \frac{3}{16} \times 8t^2 - \frac{3}{16} \times \frac{1}{12} t^4\)

\(s = \frac{1}{64} t^2 (96 - t^2)\)

(b) To find the speed when \(s = 0\), solve \(\frac{1}{64} t^2 (96 - t^2) = 0\) for \(t\):

\(t^2 = 96\), \(t = 4\sqrt{6}\).

Substitute \(t = 4\sqrt{6}\) into \(v = 16kt - \frac{1}{3} kt^3\):

\(v = \frac{3}{16} \times 16 \times 4\sqrt{6} - \frac{3}{16} \times \frac{1}{3} \times (4\sqrt{6})^3\)

\(v = 29.4 \text{ m s}^{-1}\)

(c) Maximum displacement occurs when \(v = 0\). Solve \(16kt - \frac{1}{3} kt^3 = 0\):

\(t^2 = 48\), \(t = 4\sqrt{3}\).

Substitute \(t = 4\sqrt{3}\) into \(s = \frac{1}{64} t^2 (96 - t^2)\):

\(s = \frac{1}{64} \times 48 \times (96 - 48)\)

\(s = 36 \text{ m}\)