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