Describe the motion of the bus using a speed-time graph and calculate its average speed.

The motion of the bus can be divided into four phases:

• Phase 1 (0 to 60 s): The bus moves at a constant speed of $15 \text{ ms}^{-1}$ for $1 \text{ minute}$ ($60 \text{ s}$).
• Phase 2 (60 s to 90 s): The speed increases linearly from $15 \text{ ms}^{-1}$ to $25 \text{ ms}^{-1}$ over $30 \text{ s}$.
• Phase 3 (90 s to 270 s): The bus maintains a constant speed of $25 \text{ ms}^{-1}$ for $3 \text{ minutes}$ ($180 \text{ s}$).
• Phase 4 (270 s to 300 s): The bus decelerates linearly from $25 \text{ ms}^{-1}$ to $0 \text{ ms}^{-1}$ (rest) over $30 \text{ s}$.

A speed-time graph for this motion would show:

• A horizontal line from $(0, 15)$ to $(60, 15)$, representing constant speed.
• A straight line sloping upwards from $(60, 15)$ to $(90, 25)$, representing constant acceleration.
• A horizontal line from $(90, 25)$ to $(270, 25)$, representing constant speed.
• A straight line sloping downwards from $(270, 25)$ to $(300, 0)$, representing constant deceleration.

• Phase 1: $d_1 = \text{speed} \times \text{time} = 15 \text{ ms}^{-1} \times 60 \text{ s} = 900 \text{ m}$
• Phase 2: $d_2 = \frac{(\text{initial speed} + \text{final speed})}{2} \times \text{time} = \frac{(15 \text{ ms}^{-1} + 25 \text{ ms}^{-1})}{2} \times 30 \text{ s} = 20 \text{ ms}^{-1} \times 30 \text{ s} = 600 \text{ m}$
• Phase 3: $d_3 = \text{speed} \times \text{time} = 25 \text{ ms}^{-1} \times 180 \text{ s} = 4500 \text{ m}$
• Phase 4: $d_4 = \frac{(initial speed + final speed)}{2} \times