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Gravity's effect on velocity question

  1. Oct 29, 2006 #1

    lzh

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    here is the problem:
    "a first aid kit is dropped by a rock climber who is descending steadily at 2.2m/s"
    after 2.9s, what is the velocity of the first aid kit?

    the init. velocity of the kit is 2.2 like the climber, its accerlation is 9.8 which is the accerlaeration of gravity.
    so:
    final velocity=2.2+9.8(2.9)
    which equals to 30.62
    This problem really should be easy, but for some reason my web hw service tells me that the answer is wrong.
    whats wrong?
     
  2. jcsd
  3. Oct 29, 2006 #2

    Hootenanny

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    Your equation is not quite right. Have you seen any kinematic equations before?
     
  4. Oct 29, 2006 #3

    lzh

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    but The final velocity is equal to the initial velocity plus the acceleration multiplied by the time.
    and thats what my equation is
     
  5. Oct 29, 2006 #4

    rsk

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    yes, I would have gone for v = u + at.

    I can't see yet why this isn't right. Anyone?
     
  6. Oct 29, 2006 #5

    lzh

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    Hootenany what exactly did you mean was not quite right?
     
  7. Oct 29, 2006 #6

    Hootenanny

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    Sorry, my bad. I thought we were talking about distances not velocity, I didn't read the question correctly. Have you tried entering - 30.62? Does web assign specify a number of significant figures?
     
  8. Oct 29, 2006 #7

    lzh

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    no, it accepts any answer to the 1% of the exact answer.
    i tried 30.6392 which is more exact because i used 9.80665 instead of 9.8, and it was wrong
     
  9. Oct 29, 2006 #8

    Hootenanny

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    Still have you tried entering -30.62?
     
  10. Oct 29, 2006 #9

    lzh

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    no, i have not. But wouldn't 30.6392 be a even closer number?
     
  11. Oct 29, 2006 #10

    rsk

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    i think the point is the negative sign, and it's worth a try, although the speed of the climber is given as positive.
     
  12. Oct 29, 2006 #11

    Hootenanny

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    No, since your data is only accurate to two significant figures, anything beyond that is simply 'noise', i.e. meaningless. The most number of significant figures you should quote is three, i.e. one more than your input data.
     
  13. Oct 29, 2006 #12

    Hootenanny

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    speed is always positive.
     
  14. Oct 29, 2006 #13

    lzh

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    i dont know about that...
    this question has two parts. the first part is what i posted and the second part asks for how far the kit is from the climber. I got the second part right with:
    -i figured out the kit's displacement to be 47.589 so that minus the displacement of the climber is my answer, 41.209.

    so i used that number in the kinematics forumlas that i knew to find the final velocity, and all of them resulted in 30.6...
     
  15. Oct 29, 2006 #14

    lzh

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    heres the website's description:
    a) Significant digits and precision
    The computer carries out all calculations to at least six significant digits. Do not use "significant figures" algorithms to round off your answer. Do not round off 'intermediate' calculations. Six digits are shown in solutions.
    To be scored as correct, an answer must be within 1% of the computer's answer (except for an answer of zero, which must be exact). You will be informed of any exceptions to this tolerance.
    i'm using utexas's web homework page btw.
     
  16. Oct 29, 2006 #15

    Hootenanny

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    If you are instructed not to round off then don't, enter the number as it is displayed on your calculator.
     
  17. Oct 29, 2006 #16

    rsk

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    Oops of course.

    But you know what I meant.
     
  18. Oct 29, 2006 #17

    lzh

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    ok, i'll try 30.62
     
  19. Oct 29, 2006 #18

    lzh

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    it was wrong...
     
  20. Oct 29, 2006 #19

    Hootenanny

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    I know I was being pedantic but it is usually best to clarify.

    Have you tried -30.62 yet?
     
  21. Oct 29, 2006 #20

    lzh

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    yeah, it worked.... why?
    edit: yeah speed is positive but velocity can be neg.
    but i still dont see why
     
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