(i) Using the equation of motion: \(v = u + at\), where \(v = 5\) m/s, \(u = 2\) m/s, and \(a = 0.05\) m/s², we have:

\(5 = 2 + 0.05t\)

\(0.05t = 3\)

\(t = \frac{3}{0.05} = 60\) s

Using \(v^2 = u^2 + 2as\), we find the distance \(s\):

\(5^2 = 2^2 + 2 \times 0.05 \times s\)

\(25 = 4 + 0.1s\)

\(0.1s = 21\)

\(s = \frac{21}{0.1} = 210\) m

(ii) For particle Q, the speed is given by \(v = kt^3\). The distance \(s\) is given by integrating the speed:

\(s = \int_0^{60} kt^3 \, dt = \left[ \frac{kt^4}{4} \right]_0^{60} = \frac{k \times 60^4}{4}\)

Since \(s = 210\) m, we have:

\(210 = \frac{k \times 60^4}{4}\)

\(k = \frac{210 \times 4}{60^4} = \frac{7}{108000}\)

The speed of Q at B is \(v = k \times 60^3\):

\(v = \frac{7}{108000} \times 60^3 = 14\) m/s