# A person walks up a stalled 15 m long escalator in 110 s. When standing on the same escalator,...

## Question:

A person walks up a stalled 15 m long escalator in 110 s. When standing on the same escalator, now moving, the person is carried up in 70 s. How much time would it take that person to walk up the moving escalator?

## The Frame of Reference:

In order to observe a moving body, the observer needs a frame of reference. Now frame of reference is a coordinate system, with respect to which the motion of another object is studied. The frame of reference moves as the observer moves. For a person in a moving train, everything appears to be moving past the observer. This happens because, for the observer, the frame of reference is the moving train. Therefore every motion of a moving object that is observed by the observer is with respect to the observer in his frame of reference.

Usually, for simplicity, we tend to study the problems in physics from the ground frame of reference. It should be noted that every object's motion is relative to another body. A body might appear to be at rest in one frame of reference, while it might be moving in another frame of reference. So, there can be no information about the absolute motion of an object as everything is moving with respect to something.

## Answer and Explanation:

When the man merely stands on the escalator, his speed from the ground's frame of reference is the same as the speed of escalator.

On a stalled escalator, when he walks up the escalator, his speed from the ground's frame of reference is his speed with respect to the ground. When he starts walking on a moving escalator, he attains the speed of escalator also. He was already moving with his speed. But now he also acquired the speed of escalator, thus his total speed from ground's frame of reference will be

{eq}v_{total}=v_{walk}+v_{escalator} {/eq}

When the escalator is stalled, the person walks up the escalator in 110s.

Therefore his speed of walking is

{eq}v_{walk}=\dfrac{d}{t}=\dfrac{15}{110}=0.136\:ms^{-1} {/eq}

Now when he stands still and the escalator moves, the man covers the same distance in 70s,

Therefore the speed of the escalator is

{eq}v_{esc}=\dfrac{d}{t'}=\dfrac{15}{70}=0.214\:ms^{-1} {/eq}

Thus when the man starts walking up the moving escalator, his total speed from the ground's frame of reference will be

{eq}v_{total}=0.136+0.214=0.35\:ms^{-1} {/eq}

Thus the time taken to cover the distance of escalator will be

{eq}t=\dfrac{d}{v_{total}}=\dfrac{15}{0.35}=42.82s {/eq}

#### Ask a question

Our experts can answer your tough homework and study questions.

Ask a question Ask a question#### Search Answers

#### Learn more about this topic:

from General Studies Science: Help & Review

Chapter 4 / Lesson 12