To find the velocity of A, integrate the acceleration function:

\(v(t) = \int (0.05 - 0.0002t) \, dt = 0.05t - 0.0001t^2 + C\)

Since the initial velocity is 0, \(C = 0\).

Thus, \(v(t) = 0.05t - 0.0001t^2\).

For \(t = 200\):

\(v(200) = 0.05 \times 200 - 0.0001 \times 200^2 = 10 - 4 = 6 \, \text{m/s}\)

For \(t = 500\):

\(v(500) = 0.05 \times 500 - 0.0001 \times 500^2 = 25 - 25 = 0 \, \text{m/s}\)

To find the distance traveled by A from \(t = 0\) to \(t = 500\), integrate the velocity function:

\(s = \int_0^{500} (0.05t - 0.0001t^2) \, dt\)

\(s = \left[ 0.025t^2 - \frac{0.0001t^3}{3} \right]_0^{500}\)

\(s = 0.025 \times 500^2 - \frac{0.0001 \times 500^3}{3} = 2083 \, \text{m}\)

For B, using constant acceleration and retardation:

Distance for first 200 s:

\(s_1 = \frac{1}{2} \times 6 \times 200 = 600 \, \text{m}\)

Distance for next 300 s:

\(s_2 = 6 \times 300 - \frac{1}{2} \times 0.02 \times 300^2 = 900 - 900 = 0 \, \text{m}\)

Total distance for B:

\(s_B = 600 + 900 = 1500 \, \text{m}\)

Distance between A and B at \(t = 500\):

\(2083 - 1500 = 583 \, \text{m}\)