https://file.aiquickdraw.com/mathbot/file/1730297246336xg4ghgva.jpg

To solve this problem, we need to determine the total displacement of the car from \( t = 0 \) to \( t = 6 \) seconds, based on the given acceleration-time graph. 

### Step-by-Step Solution

1. **Understanding the Problem**:
   - The acceleration graph provides the car's acceleration \( a \) over time \( t \).
   - Displacement \( s \) can be found by integrating the velocity \( v \) over time.
   - Since acceleration is the derivative of velocity, we can find the change in velocity by integrating acceleration over each time interval.

2. **Calculating Velocity**:
   - **From \( t = 0 \) to \( t = 2 \) seconds**:
     - Acceleration \( a = 4 \, \text{m/s}^2 \).
     - Change in velocity: \( \Delta v = a \times t = 4 \times 2 = 8 \, \text{m/s} \).

   - **From \( t = 2 \) to \( t = 4 \) seconds**:
     - Acceleration \( a = 0 \, \text{m/s}^2 \) (no acceleration).
     - Velocity remains constant at \( 8 \, \text{m/s} \).

   - **From \( t = 4 \) to \( t = 6 \) seconds**:
     - Acceleration \( a = -4 \, \text{m/s}^2 \).
     - Change in velocity: \( \Delta v = -4 \times 2 = -8 \, \text{m/s} \).
     - Final velocity at \( t = 6 \) seconds is \( 8 - 8 = 0 \, \text{m/s} \).

3. **Calculating Displacement**:
   - **From \( t = 0 \) to \( t = 2 \) seconds**:
     - Average velocity \( v_{\text{avg}} = \frac{0 + 8}{2} = 4 \, \text{m/s} \).
     - Displacement \( s = v_{\text{avg}} \times t = 4 \times 2 = 8 \, \text{m} \).

   - **From \( t = 2 \) to \( t = 4 \) seconds**:
     - Constant velocity \( v = 8 \, \text{m/s} \).
     - Displacement \( s = v \times t = 8 \times 2 = 16 \, \text{m} \).

   - **From \( t = 4 \) to \( t = 6 \) seconds**:
     - Average velocity \( v_{\text{avg}} = \frac{8 + 0}{2} = 4 \, \text{m/s} \).
     - Displacement \( s = v_{\text{avg}} \times t = 4 \times 2 = 8 \, \text{m} \).

4. **Total Displacement**:
   - Summing all displacements: \( 8 + 16 + 8 = 32 \, \text{m} \).

### Conclusion
The total displacement of the car from \( t = 0 \) to \( t = 6 \) seconds is \( 32 \, \text{m} \).

Since none of the given options matches 32 m exactly, it suggests a possible oversight in either the choices or calculations. Based on the analysis above, if recalculations or additional information are not required, the most accurate answer is closest to **none of the provided options**.

Would you like further clarification on any step?

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**Questions for Further Exploration**:
1. How would the displacement change if the car experienced a different pattern of acceleration?
2. What would be the effect on total displacement if the car started with an initial velocity?
3. Can you calculate the displacement using the area under the velocity-time graph?
4. How would changing the negative acceleration from \( -4 \, \text{m/s}^2 \) to \( -2 \, \text{m/s}^2 \) affect the final displacement?
5. How is the concept of integration relevant to solving displacement in this context?

**Tip**: Always double-check the area under the graph for each time interval to ensure accurate results when working with acceleration-time graphs.

The graph above shows the acceleration a of a car as a function of time t. Assuming the car starts from rest, the total displacement of the car from time t = 0 s to time t = 6 s is most nearly

Calculate the total displacement of a car from an acceleration-time graph over 6 seconds, starting from rest. This problem involves kinematics concepts like velocity, acceleration, and displacement using integration. Ideal for students in Grades 10-12.

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