A particle is moving with speed $$v = b\sqrt{x}$$ along positive x-axis. Calculate the speed of the particle at time $$t = \tau$$ (assume that the particle is at origin at t = 0)

We are told that the particle moves only along the positive x-axis and its instantaneous speed is related to its position by the relation $$v = b\sqrt{x}$$, where $$v$$ is the speed at position $$x$$ and $$b$$ is a constant with suitable units.

By definition of velocity in one dimension we have the kinematic relation

$$v \;=\; \frac{dx}{dt}$$

Substituting the given expression for $$v$$ into this definition, we obtain

$$\frac{dx}{dt} \;=\; b\sqrt{x}\, .$$

Now we separate the variables so that all terms containing $$x$$ stay on the left side and all terms containing $$t$$ remain on the right side:

$$\frac{dx}{\sqrt{x}} \;=\; b\,dt.$$

We integrate both sides. The limits for $$x$$ go from the initial position $$x = 0$$ at time $$t = 0$$ up to the general position $$x$$ at a general time $$t$$.

$$\int_{0}^{x} \frac{dx'}{\sqrt{x'}} \;=\; \int_{0}^{t} b\,dt'.$$

The integral on the left is straightforward because $$\displaystyle\int x'^{-1/2}\,dx' = 2\sqrt{x'}$$. Carrying out both integrations we get

$$2\sqrt{x}\;-\;2\sqrt{0} \;=\; b\,t\;-\;b\,(0).$$

Simplifying, $$\sqrt{x} = \dfrac{b\,t}{2}.$$ (The square-root term at the initial position is zero because the particle starts from the origin where $$x = 0$$.)

We now substitute this expression for $$\sqrt{x}$$ back into the given velocity formula $$v = b\sqrt{x}$$:

$$v \;=\; b\left(\dfrac{b\,t}{2}\right).$$

Multiplying the constants,

$$v = \dfrac{b^{2}t}{2}.$$

Finally, we are asked for the speed specifically at the instant $$t = \tau$$. Replacing $$t$$ by $$\tau$$ in the above expression yields

$$v(\tau) = \dfrac{b^{2}\tau}{2}.$$

Among the given options, this result corresponds to Option C.

Hence, the correct answer is Option C.