To find the velocity of the particle at time $t = 5$ seconds, we need to first determine the velocity function, which is the derivative of the displacement function with respect to time.

Given the displacement function: $x(t) = 2 + 6t + 2t^2$

We find the velocity $v(t)$ by taking the derivative of $x(t)$:

$v(t) = \frac{dx(t)}{dt}$

Let's calculate this derivative:

$v(t) = \frac{d}{dt} (2 + 6t + 2t^2) = 0 + 6 + 4t = 6 + 4t$

Now, substitute $t = 5$ seconds into the velocity function:

$v(5) = 6 + 4 \times 5 = 6 + 20 = 26 \, \text{m/s}$

So, the velocity of the particle at $t = 5$ seconds is $26 \, \text{m/s}$.

Would you like more details or have any questions?

Here are some related questions to consider:

1. How does the acceleration of the particle vary with time?
2. What is the initial velocity of the particle at $t = 0$?
3. At what time will the velocity be zero, if ever?
4. How far does the particle travel in the first 5 seconds?
5. What is the displacement of the particle when the velocity is $10 \, \text{m/s}$?

Tip: To find the acceleration, take the derivative of the velocity function.