To determine the time period during which the particle undergoes the greatest displacement, we need to analyze the area under the velocity-time graph for each segment. Displacement in a velocity-time graph is given by the area under the curve for that time interval.

Let's break down the time intervals between points A, B, C, and D:

1. Segment AB (0 to 4 seconds):

   • Velocity changes from 10 m/s at A to 2 m/s at B.
   • This is a trapezoidal region with area $(\frac{1}{2} \times (10 + 2) \times 4) = 24$ meters.

2. Segment BC (4 to 12 seconds):

   • Velocity remains constant at 2 m/s.
   • This is a rectangular region with area $2 \times 8 = 16$ meters.

3. Segment CD (12 to 18 seconds):

   • Velocity changes from 2 m/s at C to 14 m/s at D.
   • This is a trapezoidal region with area $(\frac{1}{2} \times (2 + 14) \times 6) = 48$ meters.

4. Conclusion:

   • The segment CD (12 to 18 seconds) has the greatest area (48 meters), representing the greatest displacement.

Thus, the particle undergoes the greatest displacement during the time period from 12 to 18 seconds.

Would you like a detailed breakdown of the calculations or have any questions?

Here are some related questions to explore further:

1. How is displacement affected if the velocity remains constant over time?
2. What does a horizontal line on a velocity-time graph represent?
3. How can you determine the acceleration from this velocity-time graph?
4. If this were a speed-time graph, would the analysis change? Why or why not?
5. How does the area under the curve relate to total distance traveled?

Tip: Remember that the area under a velocity-time graph gives the displacement, while the slope of the line gives acceleration.