Respuesta :
Answer: The speed at which the woman walks with respect to the man is 0.12 meter per second.
Explanation:
In the given problem, a man stands still on a moving escalator and a woman walks past him in the same direction as the escalator.
Both are in the same direction.
Calculate the relative speed of the woman with respect to the man.
[tex]v_{wm}=v_{w}-v_{m}[/tex]
Here, [tex]v_{wm}[/tex] is the velocity of the woman with respect to the man, [tex]v_{w}[/tex] is the velocity of the woman and [tex]v_{m}[/tex] is the velocity of the man.
Put v_{m}=0.4 meter per second and v_{w}=0.52 meter per second.
[tex]v_{wm}=0.52-0.4[/tex]
[tex]v_{wm}=0.12 ms^{-1}[/tex]
Therefore, the speed at which the woman walks with respect to the man is 0.12 meter per second.
