(i) To find the distance travelled by P in the first 3 seconds, integrate the velocity function:

\(s = \int (8t - 2t^2) \, dt = 4t^2 - \frac{2}{3}t^3 + C\)

Using the limits from 0 to 3, we have:

\(s = [4(3)^2 - \frac{2}{3}(3)^3] - [4(0)^2 - \frac{2}{3}(0)^3] = 36 - 18 = 18\) m

(ii) For t > 3, integrate the velocity function:

\(s = \int \frac{54}{t^2} \, dt = -\frac{54}{t} + C\)

Using the condition that \(s(3) = 18\), we find:

\(18 = -\frac{54}{3} + C\)

\(C = 36\)

Thus, the displacement is \(s = 36 - \frac{54}{t}\).

(iii) Set the displacement equal to 27 m:

\(36 - \frac{54}{t} = 27\)

\(\frac{54}{t} = 9\)

\(t = 6\)

Substitute \(t = 6\) into the velocity function:

\(v = \frac{54}{6^2} = \frac{54}{36} = 1.5\) m/s