(a) The acceleration \(a\) is given by differentiating the velocity function:

\(a = \frac{d}{dt}(pt^2 - qt) = 2pt - q.\)

Given that \(a = 0\) when \(t = 2\), we have:

\(2p(2) - q = 0 \Rightarrow 4p = q.\)

At \(t = 6\), the velocity must be continuous, so:

\(p(6)^2 - q(6) = 63 - 4.5(6).\)

Solving these equations:

\(36p - 6q = 36\)

\(4p = q\)

Substitute \(q = 4p\) into the first equation:

\(36p - 6(4p) = 36 \Rightarrow 36p - 24p = 36 \Rightarrow 12p = 36 \Rightarrow p = 3.\)

Then \(q = 4p = 12.\)

(b) The velocity-time graph consists of a quadratic curve from \(t = 0\) to \(t = 6\) and a straight line from \(t = 6\) to \(t = 14\). Key points are \((0, 0)\), \((6, 36)\), and \((14, 0)\).

(c) To find the total distance, integrate the velocity:

For \(0 \leq t \leq 6\):

\(s = \int_0^6 (3t^2 - 12t) \, dt = \left[ t^3 - 6t^2 \right]_0^6 = 216 - 144 = 72.\)

For \(6 \leq t \leq 14\):

\(s = \int_6^{14} (63 - 4.5t) \, dt = \left[ 63t - 2.25t^2 \right]_6^{14} = 144.\)

\(Total distance = 72 + 144 = 216 m.\)