Given the acceleration function: \(a(t) = 0.8t^{-0.75}\).

To find the velocity \(v(t)\), integrate the acceleration:

\(v(t) = \int 0.8t^{-0.75} \, dt = \frac{0.8}{0.25} t^{0.25} + C\)

\(C\) is the constant of integration. Given initial velocity \(v(0) = 1.8\), we find \(C = 1.8\).

Thus, \(v(t) = 3.2t^{0.25} + 1.8\).

To find the displacement \(s(t)\), integrate the velocity:

\(s(t) = \int (3.2t^{0.25} + 1.8) \, dt = \frac{3.2}{1.25} t^{1.25} + 1.8t + K\)

Since the particle starts from \(O\), \(s(0) = 0\), so \(K = 0\).

Thus, \(s(t) = 2.56t^{1.25} + 1.8t\).

Substitute \(t = 16\):

\(s(16) = 2.56(16)^{1.25} + 1.8(16)\)

\(s(16) = 111\)

Therefore, the displacement of \(P\) from \(O\) when \(t = 16\) is 111 m.