To determine the displacement of the object between 2.0 s and 5.0 s using the graph (assuming it is a velocity vs. time graph), you would calculate the area under the curve between 2.0 seconds and 5.0 seconds. Displacement in a velocity-time graph is given by the area under the curve for the specified time interval.

Without seeing the graph directly, here is the general process:

1. Identify the shape: If the graph forms simple geometric shapes (such as rectangles, triangles, or trapezoids), calculate the area of those shapes between the times 2.0 s and 5.0 s.
2. Calculate the area: The area corresponds to the displacement. For a constant velocity (rectangular area), the area is simply the height (velocity) times the base (time interval). For triangular or trapezoidal areas, use the appropriate area formulas.
3. Sum the areas: Add the areas of all relevant shapes between 2.0 s and 5.0 s to get the total displacement.

If you have the actual graph or more details about its shape, I could guide you through the precise calculation. Based on the given choices, one of the options is likely the result of this area calculation.

If you have a photo of the graph, feel free to upload it, and I can assist further!

Would you like to upload the graph, or do you need further clarification?

Here are 5 related questions:

1. How does the shape of a velocity-time graph affect the displacement calculation?
2. What is the significance of the area under a curve in different types of graphs?
3. How would negative velocity affect the displacement?
4. How is the average velocity between two points related to the displacement?
5. What happens to displacement when velocity is constant?

Tip: For velocity-time graphs, always focus on calculating the area between the curve and the time axis to determine displacement.