(a) To find the time for which the car is accelerating, use the formula for distance under constant acceleration:

\(s = \frac{(u+v)}{2} t\)

where \(s = 200\) m, \(u = 0\) m/s, \(v = 25\) m/s. Solving for \(t\):

\(200 = \frac{(0+25)}{2} t\)

\(200 = 12.5t\)

\(t = 16\) s

(b) The velocity-time graph is a trapezium with vertices at (0, 0), (16, 25), (32, 25), and (37, 0).

(c) The total distance traveled is the sum of the distances during acceleration, constant speed, and deceleration:

Total distance = \(200 + 400 + \frac{1}{2} \times 25 \times 5 = 662.5\) m

The total time is \(16 + 16 + 5 = 37\) s.

The average speed is:

\(\frac{662.5}{37} = 17.9\) m/s (3 s.f.)