(i) The velocity for \(0 \leq t \leq 5\) is given by \(v = 1.5 + 0.4t\). The acceleration is the derivative of velocity with respect to time, \(a = \frac{dv}{dt} = 0.4\) m s^-2.

(ii) For \(t \geq 5\), the velocity is \(v = \frac{100}{t^2} - 0.1t\). The particle is at rest when \(v = 0\). Setting \(\frac{100}{t^2} - 0.1t = 0\), we solve for \(t\):

\(\frac{100}{t^2} = 0.1t\)

\(100 = 0.1t^3\)

\(t^3 = 1000\)

\(t = 10 \text{ s}\)

(iii) For \(0 \leq t \leq 5\), the velocity is \(v = 1.5 + 0.4t\). The distance is the area under the velocity-time graph, which can be calculated using the trapezium rule:

\(\text{Distance} = \frac{1}{2} (1.5 + 3.5) \times 5 = 12.5 \text{ m}\)

For \(t \geq 5\), integrate the velocity function:

\(s(t) = \int \left( \frac{100}{t^2} - 0.1t \right) dt\)

\(= -\frac{100}{t} - 0.05t^2 + C\)

Calculate \(s(10) - s(5)\):

\(s(10) - s(5) = \left(-\frac{100}{10} - 0.05 \times 10^2\right) - \left(-\frac{100}{5} - 0.05 \times 5^2\right)\)

\(= (-10 - 5) - (-20 - 1.25)\)

\(= -15 + 21.25 = 6.25 \text{ m}\)

\(Total distance = 12.5 + 6.25 = 18.75 m.\)