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

\(\therefore V=\frac{1}{2}=0.5 \mathrm{m} / \mathrm{s}\)