To solve this problem, we can use the equations of motion and the concept of areas under a velocity-time graph.

Let the constant speed be denoted by \(V\).

1. During the first 12 seconds, the bus accelerates from rest to speed \(V\). The distance covered is given by:

\(\text{Distance} = \frac{1}{2} \times 12 \times V = 6V\)

2. During the next 30 seconds, the bus travels at constant speed \(V\). The distance covered is:

\(\text{Distance} = 30V\)

3. During the final 6 seconds, the bus decelerates to rest. The distance covered is:

\(\text{Distance} = \frac{1}{2} \times 6 \times V = 3V\)

The total distance is the sum of these distances:

\(6V + 30V + 3V = 585\)

\(39V = 585\)

\(V = \frac{585}{39} = 15 \text{ m/s}\)

Thus, the constant speed of the bus is 15 m/s.

4. To find the magnitude of the deceleration, we use the equation:

\(V = u + at\)

where \(u = 15 \text{ m/s}\), \(V = 0 \text{ m/s}\), and \(t = 6 \text{ s}\).

\(0 = 15 + a \times 6\)

\(a = -\frac{15}{6} = -2.5 \text{ m/s}^2\)

The magnitude of the deceleration is 2.5 m/s².