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Speed after body slows down with negative acceleration.

  1. Nov 28, 2007 #1
    1. The problem statement, all variables and given/known data
    Hi. I know this is a simple problem but i want to know if I am correct. Here it goes.
    Body moves with speed 12 m/s. It starts to slow down with negative acceleration 2 m/s[tex]^{}2[/tex] for 8 meters. What is his speed after those 8 meters.


    2. Relevant equations
    s=(1/2)*a*t*t
    v*v=2as
    v=v' + at
    s=v'*t + (1/2)*a*t*t

    3. The attempt at a solution
    s=(1/2)*a*t*t
    t*t=2s/a=16/2=8
    t=2,82

    v=v' - at = 12-5,65=6,34 m/s
    that's one solution the other would be.
    __________________________________
    s=v'*t - (1/2)a*t*t
    8=12*t - (1/2)*2*t*t
    t*t - 12t +8 = 0
    we get quadratic equation... we solve it and get two solutions.
    t1=0,71 s
    t2=11,29 s

    we insert this into
    v=v' - at
    v1=10,58 m/s
    v2=-10,58 m/s

    this solution is probably the correct one. But why do we get 2 values for time?
     
    Last edited: Nov 28, 2007
  2. jcsd
  3. Nov 28, 2007 #2

    rl.bhat

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    Homework Helper

    s=(1/2)*a*t*t
    t*t=2s/a=16/2=8
    t=2,82
    This formula is wrong. It should be s = ut + 1/2*at^2.
    In the problem t is not given .So you have to use v^2 = u^2 -2as. for retarding motion.
     
  4. Nov 28, 2007 #3
    Thanx. I understand.
     
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