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.