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To solve the problem of net displacement and total distance traveled by a particle with the given velocity function \( v(t) \), we need to analyze the graph provided in the image. Here's the step-by-step solution:

### Step 1: Identify the Intervals and Areas
The graph shows the velocity \( v(t) \) as a function of time \( t \). To determine the net displacement and total distance traveled, we need to find the area under the curve, considering the following:

1. Positive areas contribute positively to displacement.
2. Negative areas contribute negatively to displacement.
3. For total distance traveled, all areas contribute positively.

### Step 2: Break the Graph into Intervals
From the graph:
- Interval 1: \( t = 0 \) to \( t = 2 \)
- Interval 2: \( t = 2 \) to \( t = 4 \)
- Interval 3: \( t = 4 \) to \( t = 7 \)
- Interval 4: \( t = 7 \) to \( t = 10 \)
- Interval 5: \( t = 10 \) to \( t = 13 \)

### Step 3: Calculate the Areas
We calculate the areas of the trapezoids/triangles formed in each interval.

1. **Interval 1: \( t = 0 \) to \( t = 2 \)**
   - Area = \( \frac{1}{2} \times 2 \times 2 = 2 \, \text{m} \) (positive)

2. **Interval 2: \( t = 2 \) to \( t = 4 \)**
   - Area = \( \frac{1}{2} \times 2 \times 2 = 2 \, \text{m} \) (negative)

3. **Interval 3: \( t = 4 \) to \( t = 7 \)**
   - Area of the trapezoid = \( \frac{1}{2} \times (2 + 4) \times 3 = 9 \, \text{m} \) (positive)

4. **Interval 4: \( t = 7 \) to \( t = 10 \)**
   - Area = \( \frac{1}{2} \times (4 + 1) \times 3 = 7.5 \, \text{m} \) (negative)

5. **Interval 5: \( t = 10 \) to \( t = 13 \)**
   - Area = \( \frac{1}{2} \times 3 \times 1 = 1.5 \, \text{m} \) (positive)

### Step 4: Compute Net Displacement
Net displacement is the sum of all positive and negative areas.

\[ \text{Net displacement} = 2 - 2 + 9 - 7.5 + 1.5 = 3 \, \text{m} \]

### Step 5: Compute Total Distance Traveled
Total distance traveled is the sum of the absolute values of the areas.

\[ \text{Total distance} = 2 + 2 + 9 + 7.5 + 1.5 = 22 \, \text{m} \]

### Answers
1. **Net displacement**: 3.0000 meters
2. **Total distance traveled**: 22.0000 meters

If you have any questions or need further details, please let me know!

Here are some related questions for further practice:
1. How do you find the area under a curve in general?
2. What is the physical interpretation of the area under a velocity-time graph?
3. How does the sign of the velocity affect the net displacement?
4. How would you calculate the displacement if the graph was a curve rather than straight lines?
5. How does integrating the absolute value of velocity differ from integrating velocity itself?
6. What is the difference between displacement and distance traveled in terms of physics?
7. Can the net displacement ever be greater than the total distance traveled? Why or why not?
8. How would you calculate the total distance if the graph included horizontal sections?

**Tip:** Always consider the sign of the area when calculating net displacement, but use absolute values for total distance traveled.

Learn how to solve the problem of net displacement and total distance traveled by analyzing a velocity-time graph. This problem involves calculating areas under the curve and applying concepts of physics and calculus. Suitable for high school students.

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