Usually time and distance problems involve objects moving on fixed stationary ground base. When the base starts moving, an additional level of complexity is introduced.

## Objects move on a moving walkway

Usually time and distance problems involve objects moving on fixed stationary ground base. When the base starts moving, an additional level of complexity is introduced. These are problem areas favorable for judiciously applying very * basic relative speed concepts* and arrive at solution in a few steps.

We will elaborate basic and rich concept application in this problem area of moving platform, moving walkway or moving escalator by solving three chosen example problems.

### Problem example 1

A and B walk up an escalator one step at a time, while the escalator itself moves up at a constant speed. A walks twice as fast as B. A reaches the top in 40 steps while B reaches top in 30 steps. How many steps of the escalator can be seen when it is not moving?

- 40
- 50
- 60
- 30

### Concepts

To deal with such problems where two or more objects move at their own different speeds along a walkway that also moves at a constant speed, the simplest approach is to assume,

- The moving objects start from the starting point at the same time and deduce their distance coverage relative to each other during a suitably chosen fixed time.

This assumption makes it possible to know exactly how much distance each of the objects has moved with respect to each other during a fixed duration. It is true because distance covered by the moving walkway will be same for all objects during the time the objects cover distances relative to each other. Effectively, to know the relative distances covered, we can ignore the speed of the moving walkway.

Here we will get the relative distances covered, not the actual distances. On top of the relative distances covered, the effect of distance covered by the walkway during the time is to be applied to get the actual distances covered by the objects.

Furthermore, depending on the direction of movement of the walkway, the distance covered by the walkway is to be added or subtracted to or from the relative distances covered by the objects.

Moving walkway can be an escalator where distances will be in terms of number of steps of the escalator, while it can also be a moving sidewalk where distances will be in terms of standard distance units.

Number of objects in a moving walkway problem can be many with their speeds relative to each other given.

### Problem analysis and modelling: Given conditions

- A moves twice as fast as B.
- Escalator moves at a constant speed and the escalator, A and B all three move up, that is, move in the same direction (direction of movement could have been different).
- By taking 40 of his own steps A reaches the top of the escalator.
- By taking 30 of his own steps B reaches the top of the escalator.

### Application of Basic concepts on Given conditions towards solution: Stage 1

As A moves twice as fast as B,

- When A moves 40 of his own steps, B will move 20 of his own steps (this is true even if A and B move on a still escalator or an escalator moving at any uniform speed).
- Also B will be exactly 20 steps behind of A after A has taken 40 of his own steps and B has taken 20 of his own steps
*during the same time*(Time duration is present but it can be ignored as it is the same for all the moving objects, as well as we are concerned with only*Relative distances covered*first). - But, as by taking 40 of his own steps A reaches the top of the escalator, during the same time B will be exactly 20 steps away from the top.

Here we have used the first three given conditions that are resources to be used for solving the problem.

### Application of Basic concepts on Given conditions towards solution: Stage 2

From Stage 1,

- B is 20 steps away from the top after taking 20 of his own steps.
- By Condition 4, B reaches the top after taking 30 of his own steps.
- So to reach the top, after taking 20 of his own steps B has to take 10 more steps.
- As 20 steps of the escalator remained to be covered for reaching the top and B takes 10 of these and reaches the top, the escalator must have moved up by the remaining 10 steps during the time B has taken 10 of his own steps.
- In other words, during the time B takes 10 of his own steps the escalator also moves up by 10 steps (that is the speed of the escalator in terms of speed of Bâ€“relative again).

Finally then,

When B takes 30 of his own steps to reach the top, the escalator moves up by another 30 steps and thus helps B to reach the top.

Total number of steps is,

30+30=60.

**Answer:** Option C: 60.

### Summary mathematical reasoning

- As A moves twice as fast as B, when A takes 40 steps to reach the top, B takes 20 steps to reach exactly 20 steps behind A (because the escalator moves in the same direction as A and B, and moves at a uniform speed), that is, 20 steps away from the top.
- As B takes 30 of his own steps to reach the top, after taking 20 of his own steps, B needs to take 10 more of his own steps to reach the top, rest 10 steps having been contributed by the moving up escalator.
- So, when B takes 10 steps escalator moves up by 10 steps.
- Finally then, as B takes 30 steps to reach the top, during this time escalator would move also by 30 steps so that total number of visible steps would be, 30+30=60.

### Problem example 2

A and B walks up an escalator one step at a time while the escalator itself moves down at a constant speed. A walks twice as fast as B and reaches the top in 40 steps while B reaches top in 60 steps. How many steps of the escalator can be seen when it is still?

- 30
- 40
- 50
- 60

#### Problem solution 2

When A takes 40 steps to reach the top, B takes 20 steps and reaches a point exactly 20 steps behind A and 20 steps away from the top.

Otherwise B takes 60 steps to reach the top and so, after taking 20 steps B needs to take another 40 steps to reach the top though the top was away by only 20 steps.

This happens because by the time B takes 40 steps the escalator must have moved down by 20 steps (effectively B moves up by 20 steps to reach the top). The escalator speed is half that of B in terms of number of steps.

Given B takes 60 steps to reach the top. During this time escalator must have moved down by 30 steps. So total number of steps visible when escalator is still must be,

60 - 30=30.

**Answer:** Option A: 30.

### Problem example 3

A and B walks up an escalator one step at a time while the escalator itself moves up at a constant speed. While A takes 3 steps B takes 2 steps. A reaches the top in 60 steps while B reaches top in 50 steps. How many steps of the escalator can be seen when it is still?

- 50
- 100
- 40
- 60

#### Problem solution 3

- As escalator moves up at a constant speed, if A and B start to walk up the escalator at the same time, the time during which A takes 60 steps reaching the top, B takes two-thirds of 60, that is, 40 steps reaching a point exactly 20 steps behind A and 20 steps away from the top of the escalator.
- B reaches the top in 50 steps, but as he is 20 steps away after taking 40 steps, he still has to take 10 steps on his own to reach the top, the rest 10 steps having been aided by the upward moving escalator.
- So during the time B takes 50 steps to reach the top, escalator also moves up by 50 steps, so that total number of steps visible when escalator is standing still is, 50+50=100.

**Answer:** Option B: 100.

### End note

We have not used any formula and relied on * concept based mathematical reasoning*. The concepts used are on

*. Fastest and cleanest solution to such problems can be achieved by this approach.*

**Relative speed**### Possible problem variations

A few of the possible variations of this problem domain (problem type) are,

- Instead of two, it is possible that 3 or 4 persons move up in the same escalator. As movement of two persons is sufficient to form a complete problem where number of visible steps is to be found out, the question asked would be to find out the number steps the third person needs to take to reach the top. Relations between the speeds of the walkers would be in terms of relative speeds. Note that in escalator problems the unit of distances covered would always be number of steps.
- Instead of escalator, large number of time and distance problems can be mapped on to the moving sidewalk platform.
- Moving circular track can be a third variation.

These problem variations are a little more complex and of standard SSC CGL Tier II, CAT or GRE while the simpler versions are suitable for Part A of UGC/CSRI NET for Bioscience and Chemistry.

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