Solution:

Given the acceleration:

\[
a = 6t + 5 \, \text{m/s}^2
\]

1. Integrate acceleration to find velocity \(v\):

\[
v = \int a \, dt = \int (6t + 5) \, dt = 3t^2 + 5t + C
\]

Since it starts from rest, \(C = 0\):

\[
v = 3t^2 + 5t
\]

2. Integrate velocity to find displacement \(x\):

\[
x = \int v \, dt = \int (3t^2 + 5t) \, dt = t^3 + \frac{5}{2}t^2 + D
\]

Starting from the origin gives \(D = 0\):

\[
x = t^3 + \frac{5}{2}t^2
\]

3. Calculate displacement at \(t = 2\) s:

\[
x(2) = (2)^3 + \frac{5}{2}(2)^2 = 8 + \frac{5}{2} \cdot 4 = 8 + 10 = 18 \, \text{m}
\]

Thus, the distance covered in 2 seconds is: 18 m