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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.

Vx is the velocity of a particle moving along the x-axis as shown. If x = 2.0 m at t = 1.0 s, what is the position of the particle at t = 6.0 s?

Learn how to calculate the position of a particle at t = 6.0 seconds given its velocity-time graph and initial position. This problem uses kinematic equations and the area under the velocity-time graph to find displacement, suitable for high school students.

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