As a student of Physics, help the students in the bus (a) Graphically understand the motion of the bus up to the point when it was brought to rest. (b) Estimate the average speed of the bus for the motion. (c) Estimate the heat gained by the brake pad.

Item 3:

a) Graphically understand the motion of the bus up to the point when it was brought to rest.

Step 1: Identify the phases of motion and their corresponding velocities and times.

• Phase 1 (Constant velocity): The bus moves at $15 \text{ ms}^{-1}$ for $1 \text{ minute}$ ($60 \text{ s}$).
• Phase 2 (Acceleration): The bus accelerates from $15 \text{ ms}^{-1}$ to $25 \text{ ms}^{-1}$ in $30 \text{ s}$. This occurs from $t=60 \text{ s}$ to $t=90 \text{ s}$.
• Phase 3 (Constant velocity): The bus maintains a speed of $25 \text{ ms}^{-1}$ for $3 \text{ minutes}$ ($180 \text{ s}$). This occurs from $t=90 \text{ s}$ to $t=270 \text{ s}$.
• Phase 4 (Deceleration): The bus applies brakes and comes to rest ($0 \text{ ms}^{-1}$) in $30 \text{ s}$. This occurs from $t=270 \text{ s}$ to $t=300 \text{ s}$.

Step 2: Describe the velocity-time graph.
The velocity-time graph would show:

• A horizontal line at $v = 15 \text{ ms}^{-1}$ from $t=0 \text{ s}$ to $t=60 \text{ s}$.
• A straight line sloping upwards from $(60 \text{ s}, 15 \text{ ms}^{-1})$ to $(90 \text{ s}, 25 \text{ ms}^{-1})$.
• A horizontal line at $v = 25 \text{ ms}^{-1}$ from $t=90 \text{ s}$ to $t=270 \text{ s}$.
• A straight line sloping downwards from $(270 \text{ s}, 25 \text{ ms}^{-1})$ to $(300 \text{ s}, 0 \text{ ms}^{-1})$.

b) Estimate the average speed of the bus for the motion.

Step 1: Calculate the distance covered in each phase.

• Phase 1: $$d_1 = 15 \text{ ms}^{-1} \times 60 \text{ s} = 900 \text{ m}$$
• Phase 2: $$d_2 = \frac{15 \text{ ms}^{-1} + 25 \text{ ms}^{-1}}{2} \times 30 \text{ s} = 20 \text{ ms}^{-1} \times 30 \text{ s} = 600 \text{ m}$$
• Phase 3: $$d_3 = 25 \text{ ms}^{-1} \times 180 \text{ s} = 4500 \text{ m}$$
• Phase 4: $$d_4 = \frac{25 \text{ ms}^{-1} + 0 \text{ ms}^{-1}}{2} \times 30 \text{ s} = 12.5 \text{ ms}^{-1} \times 30 \text{ s} = 375 \text{ m}$$

Step 2: Calculate the total distance and total time.
Total distance $D = 900 \text{ m} + 600 \text{ m} + 4500 \text{ m} + 375 \text{ m} = 6375 \text{ m}$
Total time $T = 60 \text{ s} + 30 \text{ s} + 180 \text{ s} + 30 \text{ s} = 300 \text{ s}$

Step 3: Calculate the average speed.

$$\text{Average speed} = \frac{\text{Total distance}}{\text{Total time}}$$ $$$$