(a) The velocity-time graph is a trapezium with the deceleration steeper than the acceleration. The total time from start to stop is 200 s.

(b) The total distance travelled is given by the area under the velocity-time graph, which is a trapezium. The area is \(0.5 \times (170 + 200) \times V = 2775\). Solving for \(V\), we get:

\(0.5 \times 370 \times V = 2775\)

\(185V = 2775\)

\(V = 15 \text{ m s}^{-1}\)

(c) The acceleration \(a\) can be found using the formula \(a = \frac{V}{t}\), where \(t = 20 \text{ s}\). Thus:

\(a = \frac{15}{20} = 0.75 \text{ m s}^{-2}\)