Let's analyze the problem based on the image provided. The graph is a position-time graph, and we are asked to find the average velocity (in km/h) during the three segments A, B, and C of the bus's journey.

Steps to solve:

The average velocity $v_{\text{avg}}$ for each segment is calculated as:

$v_{\text{avg}} = \frac{\Delta x}{\Delta t}$

Where:

• $\Delta x$ is the change in position (displacement),
• $\Delta t$ is the change in time.

Now, let's break down each segment:

Segment A (0 to 1 hour):

• Initial position: $x_i = 40 \, \text{km}$
• Final position: $x_f = 0 \, \text{km}$
• Time interval: $t = 1.0 \, \text{hour}$

$v_{\text{avg}}(A) = \frac{x_f - x_i}{t} = \frac{0 - 40}{1.0} = -40 \, \text{km/h}$ However, the user entered $-16.67 \, \text{km/h}$, which seems incorrect, as the correct velocity should be $-40 \, \text{km/h}$.

Segment B (1 to 2 hours):

• Initial position: $x_i = 0 \, \text{km}$
• Final position: $x_f = -20 \, \text{km}$
• Time interval: $t = 1.0 \, \text{hour}$

$v_{\text{avg}}(B) = \frac{x_f - x_i}{t} = \frac{-20 - 0}{1.0} = -20 \, \text{km/h}$ The user entered $10 \, \text{km/h}$, which is also incorrect; the correct value should be $-20 \, \text{km/h}$.

Segment C (2 to 3 hours):

• Initial position: $x_i = -20 \, \text{km}$
• Final position: $x_f = 40 \, \text{km}$
• Time interval: $t = 1.0 \, \text{hour}$

$v_{\text{avg}}(C) = \frac{x_f - x_i}{t} = \frac{40 - (-20)}{1.0} = \frac{60}{1} = 60 \, \text{km/h}$ The user entered $40 \, \text{km/h}$, but the correct velocity should be $60 \, \text{km/h}$.

Summary of Correct Answers:

• (a) $v_{\text{avg}}(A) = -40 \, \text{km/h}$
• (b) $v_{\text{avg}}(B) = -20 \, \text{km/h}$
• (c) $v_{\text{avg}}(C) = 60 \, \text{km/h}$

Would you like further details on any step, or have any questions?

Here are 5 related questions you could explore:

1. How do you determine displacement from a position-time graph?
2. What is the difference between average velocity and average speed?
3. How do you calculate velocity if the graph is non-linear?
4. How can you interpret positive vs negative velocities on a graph?
5. How is instantaneous velocity different from average velocity?

Tip: Always keep in mind that velocity is a vector quantity, meaning it has both magnitude and direction.