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Thread: Dream Scenario (involving series)

  1. #1
    Feb 2009

    Dream Scenario (involving series)

    Hey everyone!

    My teacher just gave us this scenario and I can't really seem to get started on it. I was hoping someone here could lend me a hand.

    Here it is:

    " The scenario is nearly always the same. I’m standing at the end of a road in Griffith Park that I jog along each week and standing one kilometer down the road is this roadrunner bird standing there sticking his tongue out at me. I start to run after him, but I can only run in slow motion, about 1 meter per second. After one second, the road stretched uniformly and instantaneously by 1 kilometer so now the pesky roadrunner is 1998 meters away, since some of the stretch happens behind me. I try to speed up, but I’m still moving in slow motion, at 1 meter per second. After another second, the road stretches again by 1 kilometer! And this just keeps on happening. Over and over. and over, and over. Well, you get the idea. Then I wake up, hungry and frustrated.
    I have to know: Do I ever get to the roadrunner? Do I have any chance? If I get there, how long does it take? Should I take a snack to eat along the way so I don’t wake up so gosh darn hungry?
    Most of the dreams aren’t that specific. Usually, I don’t know how long the road is to begin with, or how fast I’m moving. All I know is that I’m always moving at the same slow rate, and the road stretched uniformly and instantaneously by its original amount after each second. You have to help me figure out whether or not I catch that silly roadrunner, and if so how long will it take?

    First make sure you figure out why the coyote is 1998 meters away after the first stretch.

    Next, set up a sequence {dn} where dn represents the distance between Cornelius and the coyote after n seconds but before the road does its instantaneous stretch. (e.g., d0=1000, d1=999, etc.) Then write:

    d1=1*(some expression involving d0)
    d2=2*(some expression involving d1)
    d3=3*(some expression involving d2).

    Now convert you expressions for d2 and d3 so that they only involve d0. Use this to find a general expression for dn in terms of d0.

    I've gotten a couple of things down, which may or may not be right:
    (position at time n+1) = (position at time n) + stretch - step
    d(subscript n) = position at time n
    step = 1
    stretch = (% of road in front) * 1000
    d(subscript n+1)= d(subscript n) + stretch - 1

    Any help would be greatly appreciated.

    edit: the title should say "involving sequence" not series. whoops!
    Last edited by hey light; Feb 11th 2009 at 06:51 PM.
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