To find the particle's displacement from $t = 6 \, \text{s}$ to $t = 15 \, \text{s}$, we need to calculate the area under the velocity-time graph in this interval. The area between the graph and the time axis gives the displacement, with areas above the time axis being positive and areas below being negative.

Steps:

1. From $t = 6 \, \text{s}$ to $t = 8 \, \text{s}$: The graph shows a negative velocity (below the time axis). The velocity decreases from $-5 \, \text{m/s}$ to about $0 \, \text{m/s}$. This region forms a triangle.

2. From $t = 8 \, \text{s}$ to $t = 12 \, \text{s}$: The velocity increases from $0 \, \text{m/s}$ to about $6.5 \, \text{m/s}$. This is a triangular region above the time axis.

3. From $t = 12 \, \text{s}$ to $t = 15 \, \text{s}$: The velocity decreases from $6.5 \, \text{m/s}$ to $0 \, \text{m/s}$, forming another triangle above the time axis.

I'll calculate the areas of these triangular sections to determine the net displacement.The particle's displacement between $t = 6 \, \text{s}$ and $t = 15 \, \text{s}$ is 17.75 meters.

This result was obtained by calculating the areas under the velocity-time graph, with each area corresponding to the displacement during the specified intervals.

Would you like further details or clarifications?

Here are some follow-up questions:

1. How is displacement different from distance?
2. What happens if velocity is negative during the motion?
3. How do you interpret the area above and below the time axis in a velocity-time graph?
4. How would you calculate the total distance traveled instead of displacement?
5. How does acceleration affect the shape of a velocity-time graph?

Tip: Always remember that the area under a velocity-time graph gives you displacement, and you must account for positive and negative velocities to determine net displacement.