(a) The velocity-time graph is a trapezium. The car accelerates from rest to a constant velocity, travels at this constant velocity, and then decelerates to rest. The gradient of the right-hand side is approximately twice that of the left-hand side due to the deceleration being \(2a\).

(b) To find the constant velocity, use the formula: \(v = \frac{500}{25} = 20 \text{ m s}^{-1}\).

Using the equation \(v^2 = u^2 + 2as\), where \(u = 0\), \(v = 20\), and \(s = 50\):

\(20^2 = 0 + 2a \times 50\)

\(400 = 100a\)

\(a = 4\)

(c) Time to accelerate: \(t = \frac{v}{a} = \frac{20}{4} = 5 \text{ s}\).

Deceleration time: \(t = \frac{v}{2a} = \frac{20}{8} = 2.5 \text{ s}\).

Total time = \(5 + 25 + 2.5 = 32.5 \text{ s}\).