A body of mass m starts moving from rest along x-axis so that its velocity varies as $$v = a\sqrt{s}$$ where a is a constant and s is the distance covered by the body. The total work done by all the forces acting on the body in the first t second after the start of the motion is:

The velocity at any instant is given as $$v = a\sqrt{s}$$, where $$s$$ is the distance already covered.

By definition of velocity we also have $$v = \dfrac{ds}{dt}$$.

Hence $$\dfrac{ds}{dt} = a\sqrt{s}\;.$$

We separate the variables: $$\dfrac{ds}{\sqrt{s}} = a\,dt\;.$$

Now we integrate both sides from the start of motion (when $$t = 0,\,s = 0$$) to a general instant $$t$$ (when the distance is $$s$$):

$$\int_{0}^{s}\dfrac{ds}{\sqrt{s}} = \int_{0}^{t} a\,dt\;.$$

The left integral is a standard one: the formula $$\int s^{n}\,ds = \dfrac{s^{n+1}}{n+1} + C$$ gives for $$n = -\dfrac12$$

$$\int s^{-1/2}\,ds = 2\sqrt{s}\;.$$

So we get $$\bigl[\,2\sqrt{s}\,\bigr]_{0}^{s} = 2\sqrt{s} - 0 = 2\sqrt{s}\;.$$

The right integral is simply $$a t\;.$$

Equating, $$2\sqrt{s} = a t\;.$$

Therefore $$\sqrt{s} = \dfrac{a t}{2}\quad\Longrightarrow\quad s = \dfrac{a^{2}t^{2}}{4}\;.$$

We now determine the velocity at that time. Substituting $$\sqrt{s} = \dfrac{a t}{2}$$ in the given expression $$v = a\sqrt{s}$$, we get

$$v = a\left(\dfrac{a t}{2}\right) = \dfrac{a^{2}t}{2}\;.$$

The body started from rest, so its initial kinetic energy was zero. The kinetic energy at time $$t$$ is

$$K = \dfrac12 m v^{2}\;.$$

Using $$v = \dfrac{a^{2}t}{2}$$ we write

$$K = \dfrac12 m\left(\dfrac{a^{2}t}{2}\right)^{2} = \dfrac12 m \dfrac{a^{4}t^{2}}{4} = \dfrac{m a^{4} t^{2}}{8}\;.$$

According to the work-energy theorem, the total work done by all the forces equals the change in kinetic energy:

$$W = K_{\text{final}} - K_{\text{initial}} = \dfrac{m a^{4} t^{2}}{8} - 0 = \dfrac{m a^{4} t^{2}}{8}\;.$$

Hence, the correct answer is Option 4.