And what about showing us what you've done previously ? So that we can discuss about it?
Each night after work I leave through the door D shown in the diagram and walk across the quadrangle to my car. the most direct route is obviously the north gate corridor. Half of the time this gate is locked and, as I don't have a key, I have to go across the quadrangle again towards Y, through the south gate (which is never locked) and go around the long way to my car. I used to walk directly to X, from where I can see whether the gate is locked.
a. How far on average did I walk to reach my car? Show working out.
b. Recently I have walked from the door, parallel to the wall on my left, until I can just see V. Then I can tell weather the gate is locked. When it is locked I don't have so far to walk around the long way.
How far do I walk on average under this strategy? Show working out.
c. Today I realised, when standing at Y, that I was in a straight line with X and V. I am therefore able to see from Y whether or not the gate is locked.
With this information, find a new strategy which will give the least distance on average that I walk. State this least distance and explain how it is achieved.
(Diagram is not to scale, I drew it on MS paint)
Hello Moo ^.^,
As in what I did previously on this perticular question or other questions. Honestly I haven't even start with this one yet, because currently I am studing for my next Chem. Exam. I was able just to make some time to draw that Diagram the way it looks like on the booklet*on MS paint because I have no scanner. And type all the problem out...yes by hand....for everyone to share. I have done afew of the other ones I posted though. But give me some time to post them, because I am really stressed by my next exam. Sorry Moo
What I did so far on this question is trial and error. I just find out it could be done using Pythagoras Theory, because Y X V is in are line, and cutting across in straightline is always the shortest distance. Notice the Door at the left side and the connect D with Y or X makes an triangle. Then you could calculate the length and*everything using pythagoras... I am not sure, but thats how I am doing it right now. Hope that helps.