Figure 4 is a velocity–time graph for an objectmoving with constant acceleration. Determine thedisplacement of the object over the interval 0 s to4.0 s.

To determine the displacement of the object over the interval 0 s to 4.0 s using a velocity-time graph, you need to find the area under the graph from 0 s to 4.0 s.

The area under a velocity-time graph represents the displacement of the object. If the graph is a straight line, then the area will be a triangle or a rectangle, or a combination of these two.

Since the object is moving with constant acceleration, the graph should be a straight line, indicating a constant change in velocity.

Without the specific graph, I can't provide numerical values. However, if it's a triangle, you would use the formula for the area of a triangle (1/2 * base * height). If it's a rectangle, you would use the formula for the area of a rectangle (base * height).

The base in this case would be the time interval (4.0 s - 0 s = 4.0 s), and the height would be the velocity at 4.0 s.

Add up the areas if the shape is a combination of a rectangle and a triangle.