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Finding the stoping distance.

  1. Sep 22, 2008 #1
    Acceleration = -3m/s^2
    Meters per second = 11.18

    if i were to find the stopping distance what formula should i use?

    Vo + at = Vf

    d = Vo x t + 1/2 x at^2

    Vf^2 = Vo^2 + 2ad
     
  2. jcsd
  3. Sep 22, 2008 #2
    The 11.18, is it the Vf or Vi (or both?). Don't use a formula with t because you are not given a time. I would use the last one, Vf^2-Vi^2=2ad since you have ALL that information.
     
  4. Sep 22, 2008 #3
    the 11.18 is the vi. How can i do this problem can you show me step to step? i have like 10 of these and hope to see how to do this one first.
     
  5. Sep 22, 2008 #4

    mgb_phys

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    Simple rule:
    What do you know = u, v and a (hint you know final velocity=0)
    What don't you know = t
    What are looking for = d
    which one gives you 'd' in terms of u,v,a ?
     
  6. Sep 22, 2008 #5
    using the 3rd formula i got..

    vf^2 = 125 + 2ad

    /2a both side

    vf^2 / 2a = d

    and then what else do i do?

    and is this right?
     
    Last edited: Sep 22, 2008
  7. Sep 22, 2008 #6

    mgb_phys

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    You know about how to rearrange formulae?

    vf2=vi2 + 2ad
    2ad = vf2 - vi2 and since vf=0 ie stopped
    d = - vi2 /2a

    Hint, put the units in for each number and check they cancel to give metres
     
  8. Sep 22, 2008 #7
    nvm this is what i got....

    vf^2 = 2ad

    127/2a = d

    after that?

    127a/2a

    = 127 / 2

    = 63.5?
     
  9. Sep 22, 2008 #8
    how is vf 0 when it only tells me meters per second is 11.18 and acceleration -3m/s?
     
  10. Sep 22, 2008 #9

    cepheid

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    Well, the *stopping* distance is the distance over which the object travels before *stopping*. If it is stopped, then its final velocity is ZERO. That's what it means to come to a stop...to no longer have any speed, in any direction.

    Edit: and just to complete the thought, vf is the symbol for final velocity in this problem.
     
  11. Sep 22, 2008 #10
    at the end i got d = -9 / 2a

    what do i do after this?

    -9a / 2a

    = -9 / 2

    d = -4.5

    is that right?
     
  12. Sep 22, 2008 #11

    cepheid

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    I would check your algebra if I were you. The square of the velocity is NOT equal to 9.
     
  13. Sep 22, 2008 #12
    d = 125 / 2 * -3

    d = 125 / -6


    d = -20.83

    is this right?
     
  14. Sep 22, 2008 #13

    cepheid

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    The numbers are right. The answer should not be negative because there are TWO negative signs, one on the top, and one on the bottom (the -3). So the two negative signs will cancel, if you did your algebra properly.
     
  15. Sep 22, 2008 #14

    mgb_phys

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    Remember an answer in physics without units is WRONG.
    In addition the units check that you got the formula right.
     
  16. Sep 22, 2008 #15
    mgb_phys i got -20.83. is that right?
     
  17. Sep 22, 2008 #16
    because it's like this

    d= 11.18^2 / 2 * -3

    = 125 / -6

    = -20.83

    how can it not be negative?
     
  18. Sep 22, 2008 #17

    mgb_phys

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    d = vf2 - vi2 /2a

    vi = 11.18 m/s
    vi = 0 m/s
    a = -3 m/s2

    d = 0 - 11.182 m2s-2 /(2*-3) m/s2 = 20.8 m
     
  19. Sep 22, 2008 #18

    cepheid

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    Ummm...hello? Already answered this question!

     
  20. Sep 22, 2008 #19

    LowlyPion

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    I think your original equation was wrong.

    Vf2 = Vo2 - 2*a*d

    The negative sign because your acceleration is negative (slowing)

    This then rearranges to Vo2 = 2*a*d

    No negative signs, and distance in the direction of the initial velocity.
     
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