(a) To find the velocity, integrate the acceleration function over each interval:

For \(0 \leq t < 2\), \(a = 6 + 4t\). Integrating gives \(v = 6t + 2t^2 + C\).

For \(2 \leq t < 4\), \(a = 14\). Integrating gives \(v = 14t + C\).

For \(4 \leq t \leq 16\), \(a = 16 - 2t\). Integrating gives \(v = 16t - t^2 + C\).

Using the condition \(v = 0\) at \(t = 16\), solve \(0 = 16(16) - 16^2 + C\) to find \(C = 0\).

Set \(v = 55\) and solve \(55 = 16t - t^2\) for \(4 \leq t \leq 16\). This gives \(t = 5\) and \(t = 11\).

(b) The velocity-time graph is a positive quadratic for \(0 \leq t < 2\) and a negative quadratic for \(4 \leq t \leq 16\), passing through the origin and \((16, 0)\).

(c) The distance travelled while decelerating is found by integrating the velocity from \(t = 8\) to \(t = 16\):

\(s = \int_{8}^{16} (16t - t^2) \, dt = \left[ 8t^2 - \frac{1}{3}t^3 \right]_{8}^{16}\)

\(s = \left( 2048 - 1365.33 \right) - \left( 512 - 170.67 \right) = 341 \frac{1}{3} \text{ m}\)