(i) To find \(A\), integrate the velocity function from \(t = 0\) to \(t = 15\):

\(\int_0^{15} v_1 \, dt = 225\)

\(\int_0^{15} A(t - 0.05t^2) \, dt = 225\)

\(A \left[ \frac{15^2}{2} - 0.05 \times \frac{15^3}{3} \right] = 225\)

\(A \left[ 112.5 - 0.05 \times 1125 \right] = 225\)

\(A \times 56.25 = 225\)

\(A = 4\)

To find \(B\), use the condition \(v_1(15) = v_2(15)\):

\(4(15 - 0.05 \times 15^2) = \frac{B}{15^2}\)

\(4 \times 11.25 = \frac{B}{225}\)

\(45 = \frac{B}{225}\)

\(B = 3375\)

(ii) For \(t \geq 15\), the distance travelled is:

\(s_2(t) = \int_{15}^{t} \frac{3375}{t^2} \, dt\)

\(= -3375 \left[ \frac{1}{t} - \frac{1}{15} \right]\)

\(= 225 - \frac{3375}{t}\)

Total distance: \(450 - \frac{3375}{t}\) m

(iii) Set the total distance to 315 m:

\(450 - \frac{3375}{t} = 315\)

\(\frac{3375}{t} = 135\)

\(t = 25\)

Speed at \(t = 25\):

\(v = \frac{3375}{25^2}\)

\(v = 5.4 \text{ m/s}\)