(a) To find when the acceleration is zero, differentiate the velocity function:

\(a = \frac{dv}{dt} = 3(-0.1)t^2 + 2(1.8)t - 6 = -0.3t^2 + 3.6t - 6\).

Set \(a = 0\):

\(-0.3t^2 + 3.6t - 6 = 0\).

Solving the quadratic equation gives \(t = 2\) and \(t = 10\). Thus, \(p = 2\) and \(q = 10\).

(b) The velocity at \(t = 2\) is:

\(v = -0.1(2)^3 + 1.8(2)^2 - 6(2) + 5.6 = 0\) m/s.

The velocity at \(t = 10\) is:

\(v = -0.1(10)^3 + 1.8(10)^2 - 6(10) + 5.6 = 25.6\) m/s.

The velocity-time graph is a curve with a single minimum turning point followed by a single maximum turning point.

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

\(s = \int v \, dt = \int (-0.1t^3 + 1.8t^2 - 6t + 5.6) \, dt\).

\(s = -0.025t^4 + 0.6t^3 - 3t^2 + 5.6t + C\).

Calculate the distance from \(t = 0\) to \(t = 14\):

\(s = 176.4\) m.

Calculate the distance from \(t = 14\) to \(t = 15\):

\(s = 8.025\) m.

\(Total distance = 176.4 + 8.025 = 184.425\) m.\)