(a) For particle A, it moves with constant speed 8 m/s. Since B passes O 4 seconds after A, the displacement of A after t seconds is:

\(s_A = 8(4 + t) = 32 + 8t\)

For particle B, using the equation \(s = ut + \frac{1}{2}at^2\) with \(u = 20\) m/s and \(a = -2\) m/s², the displacement is:

\(s_B = 20t - \frac{1}{2} \times 2t^2 = 20t - t^2\)

(b) Set \(s_A = s_B\):

\(32 + 8t = 20t - t^2\)

Rearrange to form a quadratic equation:

\(t^2 - 12t + 32 = 0\)

Solving this gives \(t = 4\) and \(t = 8\).

(c) The graph for A is a straight line starting at \(s = 32\) when \(t = 0\) with a positive gradient. The graph for B is an inverted quadratic starting at the origin, reaching a maximum, and then returning to the axis. They intersect at \(t = 4\) and \(t = 8\).