A body accelerates from rest at 0.2m/s2 for one-fifth of a minute - calculate the distance covered by the body.

The time given is one fifth of a minute.

$$t = \frac{1}{5} \times 60 \text{ s} = 12 \text{ s}$$

The body accelerates from rest, so the initial velocity $u = 0 \text{ m/s}$.
The acceleration $a = 0.2 \text{ m/s}^2$.
The time $t = 12 \text{ s}$.
We need to find the distance covered, $s$.
The relevant kinematic equation is:

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

$$s = (0 \text{ m/s})(12 \text{ s}) + \frac{1}{2}(0.2 \text{ m/s}^2)(12 \text{ s})^2$$ $$s = 0 + \frac{1}{2}(0.2 \text{ m/s}^2)(144 \text{ s}^2)$$ $$s = 0.1 \text{ m/s}^2 \times 144 \text{ s}^2$$ $$s = 14.4 \text{ m}$$

The distance covered by the body is $\text{14.4 m}$.