First, find the value of \(k\) by ensuring continuity of velocity at \(t = 8\):

\(30.6 - 0.9 \times 8 = \frac{1600}{8^2} + 8k\)

\(23.4 = 25 + 8k\)

\(k = -0.2\)

Next, find the time \(t\) when the particle comes to rest using \(v = \frac{1600}{t^2} + kt = 0\):

\(\frac{1600}{t^2} - 0.2t = 0\)

\(1600 = 0.2t^3\)

\(t^3 = 8000\)

\(t = 20\)

Now, integrate the velocity functions over the given intervals to find the distance:

For \(0 \leq t \leq 2\):

\(s = \int 7.2t^2 \, dt = \frac{7.2}{3}t^3 \bigg|_0^2 = 19.2 \text{ m}\)

For \(2 \leq t \leq 8\):

\(s = \int (30.6 - 0.9t) \, dt = (30.6t - \frac{0.9}{2}t^2) \bigg|_2^8 = 156.6 - 59.4 = 97.2 \text{ m}\)

For \(8 \leq t \leq 20\):

\(s = \int \left( \frac{1600}{t^2} - 0.2t \right) \, dt = \left( -\frac{1600}{t} - \frac{0.2}{2}t^2 \right) \bigg|_8^{20} = -120 - (-206.4) = 86.4 \text{ m}\)

Total distance \(OX = 19.2 + 97.2 + 86.4 = 262.2 \text{ m}\)