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A man stands still on a moving walkway and a woman walks past him in the
same direction as the escalator. To a stationary observer, the man has a
speed of 0.3 m/s and the woman has a speed of 0.38 m/s. From the frame of
reference of the man on the escalator, how fast is the woman walking?

Respuesta :

mberisso

Answer:

The woman is walking at a speed of 0.08 m/s relative to the man on the walkway.

Explanation:

Notice that for the observer, the woman is walking at 0.38 m/s, and the man (who stands still in the walkway) has a speed of 0.3 m/s.

those 0.38 m/s reported from a stationary observer are in fact the addition of two velocities: that of the walkway (0.3 m/s), and that of the woman relative to the walkway (our unknown velocity "v"):

0.38 m/s = 0.3 m/s + v

0.38 m/s - 0.3 m/s = v

v = 0.08 m/s

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