Determine the acceleration of the body down the inclined plane.

For a body sliding down a smooth inclined plane, the acceleration $a$ is given by the component of gravity along the incline.

$$a = g \sin\theta$$

Given:
Acceleration due to gravity, $g = 9.8 \text{ m/s}^2$
Angle of inclination, $\theta = 30^\circ$

Substitute the values into the acceleration formula:

$$a = (9.8 \text{ m/s}^2) \sin(30^\circ)$$ $$a = (9.8 \text{ m/s}^2) (0.5)$$ $$a = 4.9 \text{ m/s}^2$$

The body starts from rest, so the initial velocity $u = 0$.
The kinematic equation for distance $s$ is:

$$s = ut + \frac{1}{2}at^2$$

Substitute $u = 0$ and the calculated acceleration $a = 4.9 \text{ m/s}^2$:

$$s = (0)t + \frac{1}{2}(4.9 \text{ m/s}^2)t^2$$ $$s = \frac{1}{2}(4.9 \text{ m/s}^2)t^2$$ $$s = 2.45 t^2 \text{ m}$$

The distance covered after $t$ seconds is $\boxed{2.45 t^2 \text{ m}}$.