(a) To find the velocities of X and Y, integrate the acceleration functions:

For X: \\(v_x = \int (12t + 12) \, dt = 6t^2 + 12t \\\)

For Y: \\(v_y = \int (24t - 8) \, dt = 12t^2 - 8t \\\)

Set the velocities equal at collision: \\(6t^2 + 12t = 12t^2 - 8t \\\)

Solve for t: \\(6t^2 + 12t = 12t^2 - 8t \\\) leads to \\(6t^2 + 20t = 12t^2 \\\) which simplifies to \\(6t^2 = 20t \\\) and \\(t = \frac{10}{3} \\\)

Find the distance AB using displacement:

For X: \\(s_x = \int (6t^2 + 12t) \, dt = 2t^3 + 6t^2 \\\)

For Y: \\(s_y = \int (12t^2 - 8t) \, dt = 4t^3 - 4t^2 \\\)

Evaluate at \\(t = \frac{10}{3} \\\):

\(s_x = 2 \left( \frac{10}{3} \right)^3 + 6 \left( \frac{10}{3} \right)^2 = \frac{3800}{27}\)

\(s_y = 4 \left( \frac{10}{3} \right)^3 - 4 \left( \frac{10}{3} \right)^2 = \frac{2800}{27}\)

Distance AB = \\(\frac{3800}{27} - \frac{2800}{27} = \frac{1000}{27} = 37 \, \text{m} \\\)

(b) Given AB = 36 m, verify collision at \\(t = 3 \\\):

Calculate displacement for X and Y:

\(s_x = 2(3)^3 + 6(3)^2 = 54 + 54 = 108 \\\)

\(s_y = 4(3)^3 - 4(3)^2 = 108 - 36 = 72 \\\)

\(Distance AB = 108 - 72 = 36 m, verified.\)