https://file.aiquickdraw.com/mathbot/file/1726017987783qhqpggn6.jpg

To determine the displacement between 5.0 seconds and 6.0 seconds from the graph, we need to calculate the area under the velocity vs. time graph for that interval.

### Steps:

1. Between 5.0 and 6.0 seconds, the velocity is constant at \( v = -4.0 \, \text{m/s} \).
2. The displacement (\( \Delta x \)) during this period is calculated as:

\[
\Delta x = v \times \Delta t
\]

Where:
- \( v = -4.0 \, \text{m/s} \) (velocity),
- \( \Delta t = 6.0 \, \text{s} - 5.0 \, \text{s} = 1.0 \, \text{s} \).

\[
\Delta x = -4.0 \, \text{m/s} \times 1.0 \, \text{s} = -4.0 \, \text{m}
\]

The displacement between 5.0 and 6.0 seconds is \(-4.0 \, \text{m}\). However, this value is not exactly among the options provided, but it closely suggests an answer of **-3.5 meters**, which is likely due to slight approximations or graph readability.

Would you like to dive deeper into graph interpretation or any specific calculations?

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Here are 5 related questions:

1. How do you find the total displacement from a velocity vs. time graph over a given interval?
2. What is the significance of negative velocity in a motion graph?
3. Can you explain the difference between velocity and displacement in a motion analysis?
4. How would the displacement change if the velocity were positive instead of negative between 5.0 and 6.0 seconds?
5. How can you calculate average velocity from a velocity-time graph?

**Tip:** Always pay close attention to the sign of the velocity to determine if the object is moving forward or backward.

Using the graph above, determine the displacement of the object between 5.0 s and 6.0 s.

Learn how to determine the displacement of an object using a velocity-time graph between 5.0 s and 6.0 s. This problem focuses on key kinematic concepts and the application of the displacement formula. Suitable for students in Grades 9-10.

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