Expert plzzzzz give me answer

D e a r S t u d e n t, H e r e, t h e d i s \tan c e c o v e r e d b y b o t h t h e g i r l a n d e s c a l a t o r a r e t h e s a m e . L e t i t b e d . v e l o c i t y o f t h e g i r l w h o w a l k s u p t h e s t a t i o n a r y e s c a l o t o r i s, v_{g} = \frac{d i s \tan c e}{t i m e} = \frac{d}{t_{1}} V e l o c i t y o f t h e m o v i n g e s c a l a t o r i s, v_{e} = \frac{d}{t_{2}} R e s u l \tan t v e l o c i t y w h e n t h e g i r l w a l k s u p t h e m o v i n g e s c a l a t o r i s, v_{r} = v_{g} - v_{e} (B o t h a r e i n t h e s a m e d i r e c t i o n .) T i m e t a k e n b y t h e g i r l t o r e a c h t h e t o p i s, t = \frac{d i s \tan c e}{v e l o c i t y} = \frac{d}{v_{r}} \Rightarrow t = \frac{d}{\frac{d}{t_{1}} - \frac{d}{t_{2}}} = \frac{d}{d \times (\frac{1}{t_{1}} - \frac{1}{t_{2}})} = \frac{1}{(\frac{1}{t_{1}} - \frac{1}{t_{2}})} = \frac{1}{\frac{t_{2} - t_{1}}{t_{1} t_{2}}} \Rightarrow t = \frac{t_{1} t_{2}}{t_{2} - t_{1}}, O p t i o n b R e g a r d s .

View Full Answer