For upstream, Speed \(\Rightarrow v-u\)

(where \(v \rightarrow\) man and \(u \rightarrow\) water)

For downstream, Speed \(\Rightarrow v+u\)

\(t_{ up }=\frac{d}{v-u}\)

\(t_2=\frac{d}{v-u}\)

\(\Rightarrow d=(v-u) t_2 \ldots (i)\)

\(t_{ up }=\frac{d}{v-u}\)

\(t_2=\frac{d}{v-u}\)

\(\Rightarrow d=(v-u) t_2 \ldots (ii)\)

\(t_{ still }=\frac{d}{v}\)

\(t_{\text {still }}=\frac{2 t_1 t_2}{t_1+t_2}\)

On equating \((i)\) and \((ii)\)

\((v-u) t_2=(v+u) t_1\)

\(\Rightarrow v t_2-u t_2=v t_1+u t_1\)

\(\Rightarrow v\left(t_2-t_1\right)=u\left(t_1+t_2\right)\)

\(\Rightarrow u=\frac{v\left(t_2-t_1\right)}{t_2+t_1}\)

So, \(d=\left(v-\frac{v\left(t_2-t_1\right)}{t_1+t_2}\right) t_2=v t_2\left(\frac{t_1+t_2-t_2+t_1}{t_1+t_2}\right)\)

\(\frac{d}{v}=\frac{2 t_1 t_2}{t_1+t_2} \rightarrow\) Remember as shortcut