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Acceleration problems

  1. Sep 7, 2009 #1
    1. The problem statement, all variables and given/known data
    I) A car slows down from 23 m/s to rest in a distance of 80 m. What was its acceleration, assumed constant?

    II) A car slows down uniformly from a speed of 26.0 m/s to rest in 5.50 s. How far did it travel in that time?

    2. Relevant equations

    what am i doing wrooong? am i using the right formula?

    3. The attempt at a solution
    I) I used this formula: a=Vf-Vi/tf-ti but the problem doesn't give me time so it didn't work

    II) I used the same formula ^^ 0-26.0/0-5.50=4.72 is not the answer
     
  2. jcsd
  3. Sep 8, 2009 #2
    Well, you can find the time it took to travel 80m using Vavg and then find the acceleration. Or v^2 = vo^2 + 2ax
    For the second part, you need to find distance not acceleration. Which is equal to Vavg*t
     
  4. Sep 8, 2009 #3
    what's Vavg?
     
  5. Sep 8, 2009 #4

    rl.bhat

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    (v)avg = (v + vo)/2
     
  6. Sep 8, 2009 #5
    Ok, I got problem # II.

    I still don't understand # I...
    So I find the Vavg = (0+23)/2 = 11.5
    am I right here?
    Then I guess I plug that into the formula:
    11.5^2=0^2+2 a (80?) >> I don't know what's my X

    Help!!
     
  7. Sep 8, 2009 #6

    kuruman

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    Since the time is not given, use the kinematic equation that relates acceleration, displacement and the squares of the final and initial velocity.
     
  8. Sep 8, 2009 #7
    that's what I'm doing and I'm not getting the right answer
    v^2 = vo^2 + 2ax

    0=23^2+2a(80)
    0=689a
     
  9. Sep 8, 2009 #8

    Integral

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    How did you combine those factors? The first does not have an a in it so cannot simply be added.
     
  10. Sep 8, 2009 #9

    kuruman

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    Your algebra is wrong.

    0 = 232+2*a*80 is OK. It's the next step that is incorrect. Doing the number multiplications gives

    0 = 529 + 160*a

    You can't add 529 and 160*a to get 689*a !! To get 689*a you need to add 529*a + 160*a, which is not what you have here.

    Instead, you say

    160a = -529

    a = -529/160 m/s2
     
  11. Sep 8, 2009 #10
    ok,
    0(final)=23^2(initial) + 2 a (80)
    I got the 23 ^2 which is 529
    then I multiplied 80 * 2 =160
    then I added 529 +160 =689
     
  12. Sep 8, 2009 #11
    oooooooh ok,
    thanks a lot kuruman!! I really appreciate it.
     
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