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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?

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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.

What is the average velocity between the times 0 s and 0.4 s?

Learn how to calculate the average velocity between 0 s and 0.4 s from a position-time graph. This problem involves kinematic concepts and the formula for average velocity (v_avg = Δx / Δt). Suitable for high school physics students.

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