Preeti reached the metro station and found that the escalator was not working. She walked up the stationary escalator in time \(t_1\) . On other days, if she remains stationary on the moving escalator, then the escalator takes her up in time \(t_2\). The time taken by her to walk up on the moving escalator will be

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(B) \(\;\frac{{{t_1}{t_2}}}{{{t_2} + {t_1}}}\)

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\(\begin{array}{l}
Let\,{v_1}\,is\,the\,velocity\,of\,preeti\,on\,stationary\,escalator\,and\,\\
d\,is\,the\,{\rm{distance}}\,travelled\,bt\,her\\
\therefore \,{v_1} = \frac{d}{{{t_1}}}\\
Again\,let\,{v_2}\,is\,the\,velocity\,of\,escalator\,\\
\therefore \,{v_2} = \frac{d}{{{t_2}}}
\end{array}\)

\(\begin{array}{l}
\therefore \,Net\,velocity\,of\,preeti\,on\,moving\,escalator\,with\,respect\,\\
to\,the\,ground\,\\
\,\,\,\,\,\,\,\,\,\,\,v = {v_1} + {v_2} = \frac{d}{{{t_1}}} + \frac{d}{{{t_2}}} = d\left( {\frac{{{t_1} + {t_2}}}{{{t_t}{t_2}}}} \right)\\
The\,time taken by her to walk up on the moving escalatror will be\\
\,\,\,\,\,\,t = \frac{d}{v} = \frac{d}{{d\left( {\frac{{{t_1} + {t_2}}}{{{t_1}{t_2}}}} \right)}} = \frac{{{t_1}{t_2}}}{{{t_1} + {t_2}}}
\end{array}\)

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