The amplitude of wave disturbance propagating in the positive $$x$$-direction is given by $$y = \frac{1}{(1+x)^2}$$ at time $$t = 0$$ and $$y = \frac{1}{1+(x-2)^2}$$ at $$t = 1$$ s, where $$x$$ and $$y$$ are in metres. The shape of wave does not change during the propagation. The velocity of the wave will be ___ m s$$^{-1}$$.

The wave disturbance is given as $$y = \frac{1}{(1+x)^2}$$ at $$t = 0$$ and $$y = \frac{1}{1+(x-2)^2}$$ at $$t = 1$$ s.

Since the shape of the wave does not change during propagation, the wave profile at time $$t$$ can be written as $$y(x,t) = \frac{1}{(1+(x - vt))^2}$$, where $$v$$ is the wave speed in the positive $$x$$-direction.

At $$t = 0$$: $$y = \frac{1}{(1+x)^2}$$, which matches the given expression.

At $$t = 1$$ s: $$y = \frac{1}{(1 + (x - v \cdot 1))^2} = \frac{1}{(1 + (x - v))^2}$$.

Comparing with the given expression $$y = \frac{1}{1+(x-2)^2}$$, we identify $$v = 2$$ m/s.

Therefore, the velocity of the wave is $$\boxed{2}$$ m s$$^{-1}$$.