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To determine when the particle reaches \( x = 35 \, \text{m} \), we can use the information given in the velocity-time graph to find the displacement over time. The displacement \( x(t) \) can be calculated by integrating the velocity \( v_x(t) \) over the time interval of interest.

### Step-by-Step Approach:

1. **Initial position**: \( x_0 = 0 \, \text{m} \) at \( t_0 = 0 \, \text{s} \).
2. **Displacement**: The displacement at any time is given by the area under the velocity-time curve.

Let's break the graph down by time intervals:
- From \( t = 0 \, \text{s} \) to \( t = 4 \, \text{s} \), the velocity increases linearly from 0 to 10 m/s. The area under the curve (which is a triangle) gives the displacement.
- From \( t = 4 \, \text{s} \) to \( t = 8 \, \text{s} \), the velocity decreases from 10 m/s to 0 m/s. Again, the area under the curve gives the displacement.
- From \( t = 8 \, \text{s} \) to \( t = 12 \, \text{s} \), the velocity becomes negative, going from 0 m/s to -10 m/s.

I will now calculate the areas of these triangles to find the total displacement at different times and solve for when the particle is at \( x = 35 \, \text{m} \).The displacements at key times are:

- At \( t = 4 \, \text{s} \), the displacement is \( 20 \, \text{m} \).
- At \( t = 8 \, \text{s} \), the displacement is \( 40 \, \text{m} \).
- At \( t = 12 \, \text{s} \), the displacement is \( 60 \, \text{m} \).

Since we are looking for when the particle reaches \( x = 35 \, \text{m} \), this happens between \( t = 4 \, \text{s} \) and \( t = 8 \, \text{s} \), where the displacement goes from 20 m to 40 m.

Now, I'll calculate the exact time within this interval by assuming a linear relationship between time and displacement during the second section (from 4 s to 8 s).It seems like I can’t do more advanced data analysis right now. Please try again later.

However, based on the earlier analysis, the particle reaches \( x = 35 \, \text{m} \) somewhere between \( t = 4 \, \text{s} \) and \( t = 8 \, \text{s} \), and a linear interpolation would help determine the exact time.

Would you like more detailed steps on how to complete the interpolation? Here are some related questions:

1. How can I manually calculate the exact time the particle reaches \( x = 35 \, \text{m} \)?
2. What role does linear interpolation play in finding intermediate values?
3. How can you determine displacement from a velocity-time graph?
4. What does the negative velocity in the graph represent?
5. Can the particle return to the same position at different times?

**Tip**: In physics, the area under the velocity-time curve represents displacement, while the slope of a position-time graph gives velocity.

At what time or times is the particle found at x = 35 m, given the velocity-time graph and initial position x0 = 0 m at t0 = 0 s?

This problem involves finding when a particle is at 35 m by analyzing a velocity-time graph. We calculate displacement by finding the area under the curve and use linear interpolation to determine the exact time. Suitable for students in Grades 10-12.

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