1. Given relation between velocity and distance

   The body starts from rest and $v = a\sqrt{s}$

   where $a$ is a constant and $s$ is the distance covered.

2. Relate distance and time

   Since $v = \frac{ds}{dt},$ we have $\frac{ds}{dt} = a\sqrt{s}.$

   Rearranging, $\frac{ds}{\sqrt{s}} = a\,dt.$

   Integrating from $s=0$ at $t=0$ to $s=s$ at time $t$: $\int_0^s s^{-1/2} ds = \int_0^t a\,dt$ $2\sqrt{s} = at.$

   Hence, $\sqrt{s} = \frac{at}{2}.$

3. Find velocity as a function of time

   Using $v = a\sqrt{s}$, $v = a\left(\frac{at}{2}\right) = \frac{a^2 t}{2}.$

4. Use work-energy theorem

   Total work done by all forces in the first $t$ seconds is equal to the change in kinetic energy: $W = \Delta K = \frac{1}{2}mv^2 - 0.$

   Substituting $v = \frac{a^2 t}{2}$: $W = \frac{1}{2}m\left(\frac{a^2 t}{2}\right)^2$ $W = \frac{1}{2}m\cdot \frac{a^4 t^2}{4}$ $W = \frac{1}{8}ma^4 t^2.$

5. Match with options

   $W = \frac{1}{8}ma^4 t^2$

   This matches Option A.

Comparison with stored correct answer: Stored correct answer is A, which matches the derived answer.