The distance travelled is approximately equal to the area under the velocity-time graph. Estimating the areas of the trapeziums from the graph gives a total distance of approximately 45 km.
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For constant velocity, the area from 20–100 s is a rectangle: 3 × 80 = 240 m. For decreasing velocity, area from 100–120 s = ½(3+2)×20 = 50 m. Total displacement = 320 m; average acceleration = (2−0)/120 = 0.0167 ...
An object kept on the Earth can be considered at rest relative to the Earth, because its position does not change with respect to the Earth. However, relative to the Sun, it is moving.
The bus speed is 10 m s⁻¹. During reaction, it travels 5 m. Braking distance = 10²/(2 × 2.5) = 20 m. Total stopping distance = 25 m, less than 30 m. Yes, it stops safely.
During acceleration, distance = ½ × 5 × 20 = 50 m. At constant speed, distance = 20 × 10 = 200 m. During braking, distance = ½ × 6 × 20 = 60 m. Total distance = 310 m.