A bus moving at 15 m/s comes round a bend.

Step 1. Thinking distance. $15\ \text{m/s} \times 0.9\ \text{s} = 13.5\ \text{m}$.

Step 2. Braking distance. $\dfrac{(15\ \text{m/s})^2}{2 \times 4.5\ \text{m/s}^2} = 25\ \text{m}$. Stopping distance. $38.5\ \text{m}$.

Step 3. The greatest speed. $0.9v + \dfrac{v^2}{9} = 60$, so $v^2 + 8.1\ \text{m/s} \times v - 540\ \text{m}^2/\text{s}^2 = 0$, and, since $\sqrt{65.6\ \text{m}^2/\text{s}^2 + 2160\ \text{m}^2/\text{s}^2} = 47.2\ \text{m/s}$, $v = \dfrac{-8.1\ \text{m/s} + 47.2\ \text{m/s}}{2} = 19.5\ \text{m/s}$.