https://file.aiquickdraw.com/mathbot/file/1726452441807g362ywz7.jpg

To find the object's displacement at the 7.5-second mark from the velocity vs. time graph, we need to calculate the area under the velocity curve between 0 seconds and 7.5 seconds. Displacement is determined by integrating (or finding the area under) the velocity-time graph.

Let's analyze the graph:

1. From 0 to 5 seconds, the velocity is decreasing linearly from 8 m/s to -4 m/s. This section forms a trapezoid.
2. From 5 to 7.5 seconds, the velocity increases linearly from -4 m/s to 0 m/s, forming a triangle.

### Step-by-step process:

1. **From 0 to 5 seconds (Trapezoid):**
   The area of a trapezoid is given by:
   \[
   \text{Area} = \frac{1}{2} \times (v_1 + v_2) \times t
   \]
   where \(v_1 = 8 \, \text{m/s}\), \(v_2 = -4 \, \text{m/s}\), and \(t = 5 \, \text{seconds}\).

   \[
   \text{Area} = \frac{1}{2} \times (8 + (-4)) \times 5 = \frac{1}{2} \times 4 \times 5 = 10 \, \text{m}
   \]

2. **From 5 to 7.5 seconds (Triangle):**
   The area of a triangle is given by:
   \[
   \text{Area} = \frac{1}{2} \times \text{base} \times \text{height}
   \]
   Here, the base is the time interval \(2.5 \, \text{seconds}\) (from 5 to 7.5 seconds), and the height is the change in velocity, from -4 m/s to 0 m/s, which is 4 m/s.

   \[
   \text{Area} = \frac{1}{2} \times 2.5 \times 4 = 5 \, \text{m}
   \]

### Total Displacement:
Summing the areas of the two regions:

\[
\text{Total Displacement} = 10 \, \text{m} + 5 \, \text{m} = 15 \, \text{m}
\]

Thus, the displacement at 7.5 seconds is **15 meters**.

Would you like more details or have any questions?

Here are some related questions:
1. How is displacement different from distance traveled?
2. What is the physical significance of the area under a velocity-time graph?
3. How would the displacement change if the velocity were negative throughout the entire time?
4. How can you calculate displacement when the velocity function is not linear?
5. What would the total displacement be at the 15-second mark based on this graph?

**Tip:** The area under a velocity-time graph gives displacement, while the area above and below the time axis may cancel out, depending on the direction of velocity.

Use the velocity vs. time graph below to find the object's displacement at the 7.5 s mark.

Learn how to calculate the object's displacement at the 7.5-second mark using a velocity-time graph. The problem involves calculating areas under the graph using trapezoid and triangle formulas. Suitable for students in Grades 9-12.

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