https://file.aiquickdraw.com/mathbot/file/1725222410643n40sxltp.jpg

To help you with part b, I'll first analyze the content of the image.

**Analysis:**
- The image shows a velocity vs. time graph of an object's motion.
- Part b asks, "How much distance does the car cover in the first 8 seconds?"

### Steps to solve:
1. **Understand the problem:**
   - The distance covered by the car can be found by calculating the area under the velocity vs. time graph. This is because the area under the curve in a velocity-time graph represents the displacement (distance traveled in a straight line) of the object.

2. **Divide the area into shapes:**
   - The graph can be divided into rectangular and triangular areas.
   - You need to calculate the area of each shape from 0 to 8 seconds and then sum these areas to find the total distance.

3. **Calculate the area:**
   - For each rectangular section: Area = width × height.
   - For any triangular section (if applicable): Area = 0.5 × base × height.

### Calculation:

Based on the graph:
- From 0 to 2 seconds, the velocity is 3 m/s (rectangle).
- From 2 to 4 seconds, the velocity is -3 m/s (rectangle).
- From 4 to 6 seconds, the velocity is 2 m/s (rectangle).
- From 6 to 8 seconds, the velocity is -2 m/s (rectangle).

Now, let's compute the area:

1. **0 to 2 seconds:** 
   \[
   \text{Area} = 3 \, \text{m/s} \times 2 \, \text{s} = 6 \, \text{m}
   \]

2. **2 to 4 seconds:**
   \[
   \text{Area} = -3 \, \text{m/s} \times 2 \, \text{s} = -6 \, \text{m}
   \]

3. **4 to 6 seconds:**
   \[
   \text{Area} = 2 \, \text{m/s} \times 2 \, \text{s} = 4 \, \text{m}
   \]

4. **6 to 8 seconds:**
   \[
   \text{Area} = -2 \, \text{m/s} \times 2 \, \text{s} = -4 \, \text{m}
   \]

### Sum of all areas:
\[
\text{Total Distance} = 6 \, \text{m} + (-6 \, \text{m}) + 4 \, \text{m} + (-4 \, \text{m}) = 0 \, \text{m}
\]

### Conclusion:
- The distance covered by the car in the first 8 seconds is **0 meters**. This indicates that the car returns to its original position.

Would you like any further details, or do you have additional questions?

### Relative Questions:
1. How would the answer change if the graph was not symmetric?
2. What does the negative area under the curve represent in real-life scenarios?
3. How would you find the total distance if there was an acceleration involved?
4. What if the time intervals weren't uniform? How would that affect the calculation?
5. How does the concept of distance differ from displacement in this context?

### Tip:
When analyzing motion graphs, always distinguish between distance (the total path covered) and displacement (the change in position).

Learn how to calculate the distance covered by a car in the first 8 seconds using a velocity-time graph analysis. This problem involves calculating areas under the velocity-time curve, including rectangular and triangular areas. Suitable for high school students studying physics and kinematics.

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