## A Response to Terence Tao’s “An airport-inspired puzzle”

In Terence Tao’s latest post he poses three questions. Here are my solutions.

Suppose you are trying to get from one end A of a terminal to the other end B. (For simplicity, assume the terminal is a one-dimensional line segment.) Some portions of the terminal have moving walkways (in both directions); other portions do not. Your walking speed is a constant v, but while on a walkway, it is boosted by the speed u of the walkway for a net speed of v+u. (Obviously, one would only take those walkway that are going in the direction one wishes to travel in.) Your objective is to get from A to B in the shortest time possible.

- Suppose you need to pause for some period of time, say to tie your shoe. Is it more efficient to do so while on a walkway, or off the walkway? Assume the period of time required is the same in both cases.
- Suppose you have a limited amount of energy available to run and increase your speed to a higher quantity v’ (or v’+u, if you are on a walkway). Is it more efficient to run while on a walkway, or off the walkway? Assume that the energy expenditure is the same in both cases.
- Do the answers to the above questions change if one takes into account the effects of special relativity? (This is of course an academic question rather than a practical one.)
Source: Terence Tao, http://terrytao.wordpress.com/2008/12/09/an-airport-inspired-puzzle/

## Q1.

After just thinking about it without any mathematics I was not to sure so I used a mathematical approach. The first thing I did was to draw a diagram,

Admittedly, I did simplify the problem in my diagram, however I am confident that this will not affect the final answer. (How do I prove this? I don’t know.) Along with this diagram I also had to define some things in terms of variables.

As shown in the diagram, A is the starting point, B is the ending point, C is an arbitrary point in between which separates the escalator section from the non-escalator sections.

Let,

*t* = time it takes to tie shoe lace

*v* = walking speed

*u* = escalator speed

= time it takes to get from A to C

= time it takes to get from C to B

= time it takes to get from A to B

We also know, .

Now lets consider two scenarios. Scenario A, the person ties their shoe lace in the non-escalator section. Scenario B, the person ties their shoe lace in the escalator section.

### Scenario A:

### Scenario B:

Now let

I shall now make some reasonable assumptions (also formalising things a bit more),

All variables are real, and we shall assume that the person has time to tie their shoe lace while on the escalator. I.e.

I shall denote to be from scenario A and to be from scenario B. Now to see which is larger or we can examine the sign of . If it is positive then , if it is negative then .

By some algebra and as , . Hence . Therefore it would be more efficient pause for a moment while on an escalator walkway.\

## Q2.

I will take a similar approach for Q2, examining the two cases and then comparing the resultant time.

(I’ll re-edit the post when I get around to working out the solution)

q1. pause on walkway.

q2. run off walkway

q3. the same

q1:

——————————–

3 people:

A normal, B pause on walkway, C pause off walkway.

A&B: on foot, then walkway.

so A reach end first.

At that time, B need more time to reach: vT/(u+v)

A&C: walkway, then on foot.

so A reach end first.

At that time, C need more time to reach: vt/u

So, B better than C.

q2:

——————————–

3 people:

A normal, B run off walkway, C run on walkway.

A&B: walkway, then on foot.

So B reach end ealier.

At that time, A need more time to reach: dTdV/v

A&C: on foot, then walkway.

So C reach end earlier.

At that time, A need more time to reach: dTdV/(v+u)

So, B is most ealier, better than C.

q3:

——————————–

advacned theory should not be conflict with obvious analization.