We are given the velocity $V_x$ of a particle as a function of time, shown by the graph. The question asks us to find the position of the particle at $t = 6.0 \, \text{s}$, knowing that the initial position $x = 2.0 \, \text{m}$ at $t = 1.0 \, \text{s}$.

To determine the position at $t = 6.0 \, \text{s}$, we can integrate the velocity over time, as the position is the integral of velocity with respect to time:

1. From $t = 1.0 \, \text{s}$ to $t = 3.0 \, \text{s}$: The graph shows that the velocity decreases linearly from $4.0 \, \text{m/s}$ at $t = 1.0 \, \text{s}$ to $0 \, \text{m/s}$ at $t = 2.0 \, \text{s}$, and then continues to decrease to $-2.0 \, \text{m/s}$ at $t = 3.0 \, \text{s}$.

   The area under this part of the curve represents the displacement. The first part is a triangle from $t = 1.0 \, \text{s}$ to $t = 2.0 \, \text{s}$, and the second part is a triangle from $t = 2.0 \, \text{s}$ to $t = 3.0 \, \text{s}$.

2. From $t = 3.0 \, \text{s}$ to $t = 6.0 \, \text{s}$: The velocity is negative between $t = 3.0 \, \text{s}$ and $t = 5.0 \, \text{s}$, reaching zero at $t = 5.0 \, \text{s}$, and then turns positive between $t = 5.0 \, \text{s}$ and $t = 6.0 \, \text{s}$.

Let's calculate the areas under the velocity-time graph to get the displacements for each interval.

I'll calculate the total displacement from $t = 1.0 \, \text{s}$ to $t = 6.0 \, \text{s}$.The final position of the particle at $t = 6.0 \, \text{s}$ is $1.0 \, \text{m}$. Therefore, the correct answer is:

c) +1.0 m

Would you like more details or have any other questions? Here are some related questions to consider:

1. How does the area under a velocity-time graph relate to displacement?
2. What is the significance of positive and negative areas under the velocity-time graph?
3. How would the problem change if the velocity graph extended beyond $t = 6.0 \, \text{s}$?
4. What happens to the particle's motion between $t = 3.0 \, \text{s}$ and $t = 5.0 \, \text{s}$?
5. Can you derive velocity from a position-time graph in a similar way?

Tip: When dealing with velocity-time graphs, the area under the curve always represents displacement, with positive areas indicating forward motion and negative areas indicating backward motion.