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To solve the problem, we'll analyze the graph of the velocity function \( v(t) \) provided in the image. Here's a step-by-step guide to address the questions:

### 1. **Total Displacement from \( t = 0 \) to \( t = 16 \):**

Displacement is the integral of the velocity function over the given interval. We will compute the area under the curve of \( v(t) \) from \( t = 0 \) to \( t = 16 \). 

- **Between \( t = 0 \) and \( t = 4 \)**: The graph shows an increasing line, likely from \( v(0) = 0 \) to \( v(4) = 3 \).
  - Area: A triangle with a base of 4 and a height of 3.
  \[
  \text{Area}_1 = \frac{1}{2} \times 4 \times 3 = 6
  \]
  
- **Between \( t = 4 \) and \( t = 8 \)**: The graph is a triangle that decreases to \( v(8) = 0 \).
  - Area: A triangle with a base of 4 and a height of 3.
  \[
  \text{Area}_2 = \frac{1}{2} \times 4 \times 3 = 6
  \]

- **Between \( t = 8 \) and \( t = 12 \)**: The graph dips below the x-axis, indicating negative velocity, likely reaching \( v(12) = -2 \).
  - Area: A triangle with a base of 4 and a height of 2, but below the x-axis.
  \[
  \text{Area}_3 = -\frac{1}{2} \times 4 \times 2 = -4
  \]

- **Between \( t = 12 \) and \( t = 16 \)**: The graph increases slightly above the x-axis, likely reaching \( v(16) = 1 \).
  - Area: A trapezoid or broken down into a rectangle and triangle.
  \[
  \text{Area}_4 = \frac{1}{2} \times 4 \times (2 - 1) = 2 
  \]

Total Displacement:
\[
\text{Total Displacement} = 6 + 6 - 4 + 2 = 10 \text{ meters}
\]

### 2. **Total Distance Traveled from \( t = 0 \) to \( t = 16 \):**

Distance traveled considers the absolute value of the velocity:

- The areas from \( t = 0 \) to \( t = 12 \) are positive (since negative areas represent distance traveled in the opposite direction).
  - Areas from \( t = 0 \) to \( t = 8 \) are \( 6 + 6 = 12 \).
  - The area from \( t = 8 \) to \( t = 12 \) must be positive for distance, so \( |-4| = 4 \).
  - Add the distance from \( t = 12 \) to \( t = 16 \) which is \( 2 \).

Total Distance:
\[
\text{Total Distance} = 6 + 6 + 4 + 2 = 18 \text{ meters}
\]

The answers should be:
1. **Total Displacement**: \( 10 \) meters.
2. **Total Distance**: \( 18 \) meters.

Would you like further clarification or more details?

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Here are some related questions to further your understanding:

1. How do you find the displacement for a velocity function that changes over time?
2. Why do we consider the absolute value of velocity when calculating total distance?
3. How does negative velocity affect the displacement and total distance?
4. What is the difference between displacement and total distance?
5. How would you calculate the displacement if the velocity function was given as an equation instead of a graph?
6. Can you explain why the area under the velocity-time graph gives displacement?
7. How would the total displacement change if the graph dipped below the x-axis for a longer period?
8. If the particle continued to move beyond \( t = 16 \), how would you predict its displacement and distance?

**Tip:** When calculating displacement from a velocity-time graph, always account for areas above and below the x-axis separately to determine the direction of movement.

Learn how to calculate total displacement and distance traveled from a velocity graph over time. This problem involves integrating the velocity function to find displacement and using the absolute value of velocity for distance calculations. Suitable for students in Grades 11-12.

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