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Minimum distance

  1. Dec 22, 2007 #1
    1. The problem statement, all variables and given/known data
    A rectangular room measures 30 feet in length and 12 feet in height, and the ends are 12 feet in width. A fly rests at a point one foot down from the ceiling at the middle of one end and wants to get to a point one foot up from the floor at the middle of other end. Can it get there by walking less than 40 feet.


    2. Relevant equations



    3. The attempt at a solution

    I get that it needs sqrt(1658) > sqrt(1600) feet to get there. What do other people get?
     
  2. jcsd
  3. Dec 22, 2007 #2

    Shooting Star

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    Could you please show the method?
     
  4. Dec 22, 2007 #3
    Basically I unfolded the room and found a straight line along the walls from the starting point to the destination. I get that the bug walks along four different walls on this shortest path. Is it wrong?
     
  5. Dec 22, 2007 #4

    Dick

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    Pretend the room is a cardboard box and think of various ways to cut it up so it lies flat and you can draw a straight line between the two points. The shortest one I can find is sqrt(1658) as well.
     
  6. Dec 22, 2007 #5

    Shooting Star

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    I get that the bug walks along three surfaces, the one where it is, the one where it ends up, and the wall in between these two. But I get sqrt(42^2+10^2), which is even higher.

    Tell me about your four walls.
     
  7. Dec 22, 2007 #6

    Dick

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    There's a shorter path crossing four surfaces. Angle up to the ceiling, go along the ceiling for a while, cross on to the side wall and then move on to the end wall. Get those scissors out.
     
  8. Dec 22, 2007 #7

    Shooting Star

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    As far as I can tell, the answer to the original question is no. But I'm waiting for a better solution.

    Incited :mad: by Dick's snippy post (well, he was talking about scissors :wink:), I have destroyed a few rooms and ultimately found the shortest path as 40 ft. This comes from sqrt[(1+30+1)^2 + (12+12)^2] = 40. I'm sure you guys can visualize the way I've opened up the room, without giving any thought to the fact I'm scared of shooting stars falling from the sky.

    What about < 40 ft?
     
  9. Dec 22, 2007 #8

    Dick

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    Oooops. The way I was thinking was sort of ceiling obsessive. Go to a side wall, traverse the wall, then back through the end wall. sqrt(10^2+36^2). It IS less than 40! Thanks for pushing me into rethinking this. I'm still not sure I see the 40 route.
     
    Last edited: Dec 22, 2007
  10. Dec 22, 2007 #9

    Gib Z

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    Alternative solution;

    Yes, by flying there instead =D
     
  11. Dec 23, 2007 #10
    Hmmmm the only way I see is by flying. The shortest path I get is 42 exactly.
     
  12. Dec 23, 2007 #11

    Dick

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    Read the question. The fly has to walk. If it can fly the shortest distance is sqrt(10^2+30^2). Which is definitely not 42.
     
    Last edited: Dec 23, 2007
  13. Dec 23, 2007 #12
    I read it. It doesn't say the fly has to walk. And I dont see how you can get sqrt(10^2+30^2).
     
  14. Dec 23, 2007 #13

    Gib Z

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    I was actually joking when I last posted in this thread >.< Even if it didn't explicitly state, I assumed it had to walk :(
     
  15. Dec 23, 2007 #14
    Well Gib, I don't see why they would bother making it a fly instead of.. say... a spider. Seems like they are inviting ambiguity. Or.. maybe they did it on purpose and flying is the correct answer.
     
  16. Dec 23, 2007 #15
    Far as I can see.. there are four paths the fly could take (if walking): If it crosses the ceiling or floor, the distance is 1+30+11=42. If it crosses one of the two side walls, the distance is sqrt(42^2 + 10^2) which is even longer.
     
  17. Dec 23, 2007 #16

    Shooting Star

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    I'm very bad at making picture files, and I don't have a scanner, so I can't show you the path right now. I'll try -- maybe ask someone to make a bmp or whatever file.

    But could you show us your path? Or describe in detail?
     
  18. Dec 23, 2007 #17
    you nearly got it! you should have gone a little further.

    you should unfold to two dimension in such a way that you draw the small sidewalls up-left and down-right, and (from up to down) the ceiling, one big sidewall and the ground in the middle.

    Now draw a straight line. it's 40 ft.

    You could also try take 6 walls, but as you can see that would only be longer. I think 40 ft. is the minimum.
     
    Last edited: Dec 23, 2007
  19. Dec 23, 2007 #18
    oh wait someone said this already >_> never mind then
     
  20. Dec 23, 2007 #19

    Shooting Star

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    The 40 ft route may be possible to see even without the diagram because of the way I've written it: sqrt[(1+30+1)^2 + (12+12)^2].

    But where are you gettting the 10 from? Please confirm once more.
     
  21. Dec 23, 2007 #20
    In the book where I found the problem (problem solving through problems) the fly has a broken wing and does explicitly have to walk. I should have written that.
     
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