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To determine the object's position at \( t = 8 \) seconds based on the given velocity-time graph, we need to calculate the object's displacement. Displacement is the integral (or the area under the curve) of the velocity function with respect to time. The graph shows velocity vs. time, so we will calculate the area between the curve and the time axis for each segment of the graph, ensuring we consider both positive and negative areas.

### Breakdown of the graph:

1. **From \( t = 0 \) to \( t = 2 \):** 
   - The velocity is increasing linearly from 0 m/s to 2 m/s, forming a triangle.
   - Area of triangle: 
     \[
     A_1 = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 2 \times 2 = 2 \, \text{m}
     \]

2. **From \( t = 2 \) to \( t = 4 \):** 
   - The velocity decreases linearly from 2 m/s to 0 m/s, again forming a triangle.
   - Area of triangle: 
     \[
     A_2 = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 2 \times 2 = 2 \, \text{m}
     \]

3. **From \( t = 4 \) to \( t = 6 \):** 
   - The velocity decreases linearly from 0 m/s to -1 m/s, forming a triangle below the time axis (negative velocity means negative displacement).
   - Area of triangle:
     \[
     A_3 = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 2 \times 1 = 1 \, \text{m}
     \]
   - This is a negative area: \( A_3 = -1 \, \text{m} \).

4. **From \( t = 6 \) to \( t = 8 \):**
   - The velocity remains constant at \( -1 \) m/s, forming a rectangle.
   - Area of rectangle:
     \[
     A_4 = \text{base} \times \text{height} = 2 \times (-1) = -2 \, \text{m}
     \]

### Total displacement:

\[
\text{Total displacement} = A_1 + A_2 + A_3 + A_4 = 2 + 2 - 1 - 2 = 1 \, \text{m}
\]

Thus, at \( t = 8 \) seconds, the object is 1 meter from the origin. The positive displacement means it is 1 meter in the positive direction.

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Would you like further details or have any questions?

Here are some related questions for you to explore:

1. What is the total distance traveled by the object up to \( t = 8 \) seconds?
2. What would the object's displacement be at \( t = 6 \) seconds?
3. How would you calculate the object's average velocity over the entire 8-second period?
4. What is the difference between displacement and total distance traveled?
5. How would the position change if the object started at 5 meters instead of the origin?

**Tip:** Remember that displacement accounts for direction, while total distance traveled is the sum of the absolute values of the areas under the velocity-time curve.

The graph below shows the velocity of an object over time. If this object starts at the origin at time 0, where is it at time t = 8 s? Explain how you reached your answer.

Learn how to calculate the displacement of an object using a velocity-time graph. This problem involves finding the areas under the graph for different intervals and considering both positive and negative displacements. Suitable for students in Grades 9-12.

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