(a) The car accelerates from 0 to 20 m s^-1 at a rate of 2 m s^-2. Using the formula for acceleration, \(v = u + at\), where \(u = 0\), \(v = 20\), and \(a = 2\), we have:

\(20 = 0 + 2T\)

\(T = \frac{20}{2} = 10\)

(b) The distance travelled before the constant speed is the area under the graph up to that point. This includes a triangle and a trapezium:

Distance before constant speed = \(\frac{1}{2} \times 10 \times 20 + \frac{1}{2} \times (20 + V) \times 5\)

= \(100 + \frac{1}{2} \times (20 + V) \times 5\)

= \(100 + 2.5(20 + V)\)

= \(150 + 2.5V\)

The distance travelled after the constant speed is:

Distance after constant speed = \(27.5V + \frac{1}{2} \times V \times 5\)

= \(27.5V + 2.5V\)

= \(30V\)

Given that the distance before constant speed is one third of the total distance:

\(150 + 2.5V = \frac{1}{3}(150 + 2.5V + 30V)\)

\(150 + 2.5V = \frac{1}{3}(150 + 32.5V)\)

\(450 + 7.5V = 150 + 32.5V\)

\(300 = 25V\)

\(V = 12\)