a person walks up a statiomary elavator in t1 sec .If he is standing on elavator and then the elavator carries him up in t2 sec . the length of elavator is L . Then caluclate velocity of elavator ,velocity of mam and if man starts to walk on moving elavator in the direction of motion of elavator ,then time taken by the man to move up is please solve i will make as brainlist
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Answer:
a. As the escalator is stationary, so the distance covered in t
1
seconds is L which is the length of the escalator.
Speed of man w.r.t. the escalator, v
me
=
t
1
L
b.When the man is stationary, by taking man as a reference point the distance covered by escalator is L in time t
2
.
Speed of escalator, v
c
=
t
2
L
c. Speed of man w.r.t. the ground, v
m
=v
me
+v
c
_
⇒v
m
=
t
1
L
+
t
2
L
=L[
t
1
1
+
t
1
1
]= L[
t
1
t
2
t
1
+t
2
]
∴t=
v
m
L
=[
t
1
+t
2
t
1
t
2
]
which is the time taken by the man to walk up the moving escalator.
Explanation:
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