obtained by putting $X=x-vt$
$y=\frac{1}{1+x^{2}}=\frac{1}{1+(x-v t)^{2}}$ $...(i)$
We know at $t=2 \mathrm{sec}$
$y=\frac{1}{1+(x-1)^{2}}$ $...(ii)$
On comparing$ (i)$ and $(ii)$ we get
$v t=1$
$V=\frac{1}{t}$
As $t=2$ sec
$\therefore V=\frac{1}{2}=0.5 \mathrm{m} / \mathrm{s}$