The graph in the image shows the velocity versus time for an object from $t = 0 \, \text{s}$ to $t = 10 \, \text{s}$. We need to calculate the total distance traveled by the object during this time interval.

To find the total distance, we calculate the absolute area under the velocity-time curve, as the area under the curve represents the displacement, and the total distance is the sum of the absolute values of these areas.

Analysis:

1. From $t = 0 \, \text{s}$ to $t = 2 \, \text{s}$:

   • Velocity: $2 \, \text{m/s}$
   • The area is a rectangle: $\text{Area} = 2 \times 2 = 4 \, \text{m}$.

2. From $t = 2 \, \text{s}$ to $t = 8 \, \text{s}$:

   • The velocity is negative (the object is moving in the opposite direction).
   • The area is a triangle with a base of $6 \, \text{s}$ and a height of $-4 \, \text{m/s}$.
   • The area of the triangle: $\text{Area} = \frac{1}{2} \times 6 \times 4 = 12 \, \text{m}$.

3. From $t = 8 \, \text{s}$ to $t = 10 \, \text{s}$:

   • Velocity: $2 \, \text{m/s}$
   • The area is a rectangle: $\text{Area} = 2 \times 2 = 4 \, \text{m}$.

Total Distance:

The total distance is the sum of the absolute values of the areas: $\text{Total Distance} = 4 + 12 + 4 = 20 \, \text{m}$

Thus, the total distance traveled by the object during this time interval is 20 meters.

If you have any questions or need more details, let me know! Here are some related questions for further understanding:

1. How does the total distance differ from displacement?
2. What is the significance of the area under the velocity-time graph?
3. How would the calculation change if the graph had a curve instead of straight lines?
4. Can you explain why we take the absolute value of the area to calculate distance?
5. What would happen to the total distance if the object were to move back to its original position?

Tip: Always ensure that when calculating total distance, the direction of motion doesn't matter, so you sum the absolute values of the areas under the velocity-time graph.