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

0.52 - 0.4 = 0.12
0.12 m/s

Answer 2

Answer: The speed at which the woman walks with respect to the man is 0.12 meter per second.

Step-by-step 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.