To solve for the average velocity between the times 0 s and 0.4 s, we use the formula for average velocity:

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

where:

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

From the graph:

• At $t = 0 \, \text{s}$, the position is $x = 0 \, \text{m}$,
• At $t = 0.4 \, \text{s}$, the position is $x = 4 \, \text{m}$.

So, the change in position $\Delta x = 4 - 0 = 4 \, \text{m}$ and the change in time $\Delta t = 0.4 - 0 = 0.4 \, \text{s}$.

Now, substituting into the formula for average velocity:

$v_{\text{avg}} = \frac{4 \, \text{m}}{0.4 \, \text{s}} = 10 \, \text{m/s}$

Thus, the average velocity between 0 s and 0.4 s is $10 \, \text{m/s}$.

Would you like further details or explanations?

Here are 5 related questions to expand your understanding:

1. How would the average velocity change if the time interval was from 0.4 s to 0.8 s?
2. What is the significance of the slope of the position-time graph?
3. How can you determine if the object is accelerating from the graph?
4. What does a negative slope in the position-time graph indicate about the velocity?
5. How does the average velocity differ from instantaneous velocity?

Tip: Always ensure you check the units when calculating velocity, position, or time to avoid errors in interpretation.