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Car motion

  1. Mar 27, 2009 #1
    Car A accelerates at 4.2ms
    Car B accelerates at 3.5ms
    In a drag Car B leaves the start line .80 secs earlier than car A.
    At what time does car A pass car B?
    My attempt:
    Car A:
    y=4.2mx

    Determine constant for car B
    v=u+at
    v=0+(3.5)(.80)=2.8
    s=1/2(0+2.8).80
    s=1.12metres

    Car B:
    y=3.5x + 1.12

    3.5x + 1.12 = 4.2x
    to find when graph intercepts
    -0.7x=-1.12
    x=1.6 seconds
    What am I doing wrong?
     
  2. jcsd
  3. Mar 27, 2009 #2

    LowlyPion

    User Avatar
    Homework Helper

    For one thing you have apparently forgotten the acceleration term.
    For the other you have forgotten what your initial velocity was.

    Xb = 1.12 + 2.8t + 1/2*3.5*t2

    Xa = 1/2*4.2*t2

    When the distances are the same then t is your time.

    Of course it will require a quadratic solution.
     
  4. Mar 29, 2009 #3
    why is the squared variable multiplied by 1/2?
     
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