(i) To find \(k_1\), use the formula for distance: \(s = \int v \, dt\). For \(0 \leq t \leq 60\),

\(s = \int_0^{60} (k_1 t - 0.005t^2) \, dt = \left[ \frac{k_1 t^2}{2} - 0.005 \frac{t^3}{3} \right]_0^{60} = 540\)

\(k_1 \frac{60^2}{2} - 0.005 \frac{60^3}{3} = 540\)

\(1800k_1 - 72 = 540\)

\(1800k_1 = 612\)

\(k_1 = 0.5\)

For \(k_2\), equate \(v(60)\) from both expressions:

\(v(60) = 0.5 \times 60 - 0.005 \times 60^2 = \frac{k_2}{\sqrt{60}}\)

\(30 - 18 = \frac{k_2}{\sqrt{60}}\)

\(12 = \frac{k_2}{\sqrt{60}}\)

\(k_2 = 12\sqrt{60}\)

(ii) For \(t \geq 60\), the distance is:

\(s = 540 + \int_{60}^{t} \frac{12\sqrt{60}}{\sqrt{t}} \, dt\)

\(s = 540 + 12\sqrt{60} \left[ 2\sqrt{t} - 2\sqrt{60} \right]\)

\(s = 24\sqrt{60t} - 900\)

(iii) Set \(s(t) = 1260\):

\(24\sqrt{60t} - 900 = 1260\)

\(24\sqrt{60t} = 2160\)

\(\sqrt{60t} = 90\)

\(60t = 8100\)

\(t = 135\)

Find speed at \(t = 135\):

\(v = \frac{12\sqrt{60}}{\sqrt{135}} = 8 \text{ m s}^{-1}\)