Question

A man stands still on a moving escalator and a woman walks past him in the same direction as the escalator. To a stationary observer, the man has a speed of 0.4 m/s and the woman has a speed of 0.52 m/s. From the frame of reference of the man on the escalator, how fast is the woman walking?

Answer 1

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.

$v_{wm}=v_{w}-v_{m}$

Here, $v_{wm}$ is the velocity of the woman with respect to the man, $v_{w}$ is the velocity of the woman and $v_{m}$ is the velocity of the man.

Put v_{m}=0.4 meter per second and v_{w}=0.52 meter per second.

$v_{wm}=0.52-0.4$

$v_{wm}=0.12 ms^{-1}$

Therefore, the speed at which the woman walks with respect to the man is 0.12 meter per second.  

             

Answer 2

0.52 - 0.4 = 0.12
0.12 m/s