Question 1189589
Here's how to solve this problem:

1. **Find the displacement function:**

The velocity function V(t) is the derivative of the position function s(t). To find the position function, we need to integrate the velocity function:

s(t) = ∫V(t) dt = ∫√(3t - 1) dt

Let u = 3t - 1, so du = 3 dt, and dt = du/3.  Substituting:

s(t) = ∫√u * (du/3) = (1/3) ∫u^(1/2) du
s(t) = (1/3) * (2/3) * u^(3/2) + C = (2/9) * (3t - 1)^(3/2) + C

where C is the constant of integration.

2. **Solve for the constant of integration (C):**

We are given that at t = 2, the particle's position is 8 meters.  Plug in these values to solve for C:

8 = (2/9) * (3*2 - 1)^(3/2) + C
8 = (2/9) * 5^(3/2) + C
8 = (2/9) * 5√5 + C
C = 8 - (10√5)/9

3. **Find the position at t = 7:**

Now that we have the complete position function, we can find the position at t = 7:

s(7) = (2/9) * (3*7 - 1)^(3/2) + 8 - (10√5)/9
s(7) = (2/9) * 20^(3/2) + 8 - (10√5)/9
s(7) = (2/9) * 20√20 + 8 - (10√5)/9
s(7) = (40√5)/9 + 8 - (10√5)/9
s(7) = (30√5)/9 + 8
s(7) = (10√5)/3 + 8
s(7) ≈ (10 * 2.236) / 3 + 8
s(7) ≈ 7.453 + 8
s(7) ≈ 15.453

Therefore, the particle's position at t = 7 seconds is approximately 15.45 meters.