(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:

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

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

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

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 \(t = 500\):

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

(i) The speed of P at A is given by the formula for final velocity:

\(v = u + at\)

where \(u = 1.3\) m s^-1, \(a = 0.1\) m s^-2, and \(t = 20\) s.

\(v = 1.3 + 0.1 \times 20 = 3.3\) m s^-1

The speed of Q at B is the same as the speed of P at A, so \(v_Q(20) = 3.3\) m s^-1.

The acceleration of Q is \(0.016t\), so the velocity function is:

\(v_Q(t) = \int 0.016t \, dt = 0.008t^2 + v_Q(0)\)

Substituting \(t = 20\) and \(v_Q(20) = 3.3\):

\(3.3 = 0.008 \times 20^2 + v_Q(0)\)

\(3.3 = 3.2 + v_Q(0)\)

\(v_Q(0) = 0.1\) m s^-1

(ii) The distance AO is calculated using the formula for distance under constant acceleration:

\(s = ut + \frac{1}{2}at^2\)

\(AO = 1.3 \times 20 + \frac{1}{2} \times 0.1 \times 20^2 = 46\) m

The distance OB is calculated using the integral of velocity:

\(OB = \int (0.008t^2 + 0.1) \, dt\) from 0 to 20

\(OB = \left[ \frac{0.008}{3}t^3 + 0.1t \right]_0^{20}\)

\(OB = \frac{0.008}{3} \times 20^3 + 0.1 \times 20 = 23.3\) m

Therefore, the total distance \(AB = AO + OB = 46 + 23.3 = 69.3\) m

To find the time T when particles P and Q collide, we set up equations for their motions.

For particle P, starting from rest with acceleration 0.5 m s^-2, the distance covered is given by:

\(s_{AP} = \frac{1}{2} a t^2 = \frac{1}{2} \times 0.5 \times T^2 = 0.25 T^2\)

For particle Q, moving with constant speed 0.75 m s^-1, the distance covered is:

\(s_{BQ} = 0.75 T\)

Since the total distance between A and B is 10 m, we have:

\(s_{AP} + s_{BQ} = 10\)

Substituting the expressions for \(s_{AP}\) and \(s_{BQ}\), we get:

\(0.25 T^2 + 0.75 T = 10\)

This simplifies to the quadratic equation:

\(0.25 T^2 + 0.75 T - 10 = 0\)

Multiplying through by 4 to clear the fraction:

\(T^2 + 3T - 40 = 0\)

Factoring gives:

\((T + 8)(T - 5) = 0\)

Thus, \(T = 5\) (ignoring the negative solution \(T = -8\)).

For the speed of P immediately before the collision, we use:

\(v = u + at\)

Since \(u = 0\) and \(a = 0.5\), we have:

\(v = 0 + 0.5 \times 5 = 2.5 \text{ m s}^{-1}\)

(i) For cyclist P, using the equation of motion:

\(s = ut + \frac{1}{2}at^2\)

where \(u = 3\) m s^-1, \(a = 0.005\) m s^-2, and \(t = 400\) s:

\(AB = 3 \times 400 + \frac{1}{2} \times 0.005 \times 400^2 = 1600 \text{ m}\)

To find P's speed at B:

\(v_B = u + at = 3 + 0.005 \times 400 = 5 \text{ m s}^{-1}\)

(ii) For cyclist Q, integrate the velocity function to find the distance:

\(\int_0^{400} v \, dt = 1600\)

\(\int_0^{400} (0.04t - 0.0001t^2 + k) \, dt = 1600\)

\([0.02t^2 - 0.0001 \frac{t^3}{3} + kt]_0^{400} = 1600\)

\(400k = 1600 - 0.02 \times 400^2 + 0.0001 \times \frac{400^3}{3}\)

\(k = 4 - \frac{8 + 16/3}{3} = \frac{4}{3}\)

To find the maximum speed, differentiate \(v\) and set \(\frac{dv}{dt} = 0\):

\(\frac{dv}{dt} = 0.04 - 0.0002t = 0\)

\(t = 200\) s, \(v_{\text{max}} = 0.04 \times 200 - 0.0001 \times 200^2 + \frac{4}{3} = 5.33 \text{ m s}^{-1}\)

(iii) For cyclist P, time taken from B to C:

\(\text{Time} = \frac{1400}{5} = 280 \text{ s}\)

For cyclist Q, using \(s = ut + \frac{1}{2}at^2\):

\(1400 = \frac{4}{3} \times 280 + \frac{1}{2} \times a \times 280^2\)

\(a = 0.0262 \text{ m s}^{-2}\)