Question

Escalator question

Answer 1

k

Answer 2

here is my reasoning:
let's suppose that the man is moving with a speed \(v\) with respect to the escalator
and the speed of the escalator is \(V\)

Answer 3

time needed to the man in order to go upward, is:
\[\Large {t_1} = \frac{d}{{v + V}}\]
where time needed to the man in order to go dowward, is:
\[\Large {t_2} = \frac{d}{{v - V}}\]
where \(d\) is the vertical distance

Answer 4

now, combining such equations, we get:
\[\Large v = \frac{{{t_2} + {t_1}}}{{{t_2} - {t_1}}}V = \frac{{90 + 30}}{{90 - 30}}V\]

Answer 5

yes! It is \(t_2=90\) and \(t_1=30\)

Answer 6

how can i find a.) from here

Answer 7

from the formulas above, we get:
\(v=2V\)

Answer 8

yes

Answer 9

when the escalator is stationary, we have \(V=0\), since the escalator is not working

Answer 10

do we know the speed of the escalator?

Answer 11

yes

Answer 12

please what is such speed?

Answer 13

ya wait

Answer 14

sry i think i don't have

Answer 15

do you have numerical options?

Answer 16

here in the link in 19:17 min

the guy explains it , but i didnt understand

https://www.youtube.com/watch?v=ByRhu23WNTY

Answer 17

I think that we have too many unknowns

Answer 18

I'm sorry I don't know how to go on

Answer 19

when the man goes up, he takes 30 seconds, and the distance covered is a multiple of \(30 \times (v+V)=30 \times 3v=90v\), so by substitution, the time needed to go up when escalator is not working, is:
\[\Large t = \frac{d}{v} = \frac{{90v}}{v} = 45\sec \]
the teacher of the video, consider a factor of proportionality \(x\), nevertheless we don't neeed of such proportionality factor, since the speed of the man which goes up is proportional to \(v\) and the space to go up is proportional to \(30(v+V)=30(v+2v)=90v\)

Answer 20

similarly, when escalator is working, I think that the time needed to the man in order to go up, is:
\[\Large {t_1} = \frac{d}{{v + V}} = \frac{{90v}}{{2v}} = 45\sec \]
since the man is still, so its speed \(v=0\)

Answer 21

sorry I have made a typo;
\[\Large {t_1} = \frac{d}{{v + V}} = \frac{{90v}}{{v/2}} = 180\sec \]
since \(V=v/2\)

Answer 22

\[\frac{d}{v-V} = 90\]
\[v = 2V\]
\[\rightarrow d = 90V = 45v\]
Therefore when V = 0 (escalator not working) the time is:
\[t = \frac{d}{v} = \frac{45v}{v} = 45 s\]
Similarly, when v = 0 (man rides escalator) the time is:
\[t = \frac{d}{V} = \frac{90V}{V} = 90 s\]

Answer 23

yes! correct! @dumbcow
sorry @mathmath333 my second result is wrong