The velocity of the particle is given by the function:

$v(t) = \frac{5t^2}{t^2 + 81}$

To find the distance $d$ traveled by the particle after $t = 2$ seconds, we need to compute the displacement, which is the integral of the velocity function over the interval from $t = 0$ to $t = 2$:

$d = \int_0^2 v(t) \, dt = \int_0^2 \frac{5t^2}{t^2 + 81} \, dt$

Let's compute this integral.

Step-by-step solution:

1. Setup the integral: $d = \int_0^2 \frac{5t^2}{t^2 + 81} \, dt$

2. Solve the integral using substitution and standard integral techniques.

I'll calculate this now.The result of the integral is:

$d = 10 - 45 \cdot \arctan\left(\frac{2}{9}\right)$

Now, let's calculate the numerical value of this expression to determine the distance traveled after $t = 2$ seconds.The distance traveled by the particle after 2 seconds is approximately $d = 0.16$ meters.

Would you like more details on the calculation, or do you have any questions?

Here are some related questions you might be interested in:

1. How would the distance change if the time interval was from 0 to 3 seconds?
2. What would happen to the velocity function if the constant 81 in the denominator were changed?
3. How can we interpret the arctangent term in the result?
4. What is the physical significance of integrating the velocity function?
5. Can we find the particle's maximum velocity?

Tip: Integrating a velocity function gives the total displacement, while integrating the absolute value of velocity provides total distance traveled.