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Car accelerating uniformly for 6s - kinematics

  1. Sep 4, 2006 #1
    A car starts from rest and accelerates uniformly for 6s. The distance traveled in the 6s is greater than the distance traveled in the 1st second by a factor of what?

    how would i go about doing this?
     
  2. jcsd
  3. Sep 4, 2006 #2

    quasar987

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    Use the equation of kinematics at constant acceleration:

    [tex]x(t)=x_0+v_0t+\frac{a}{2}t^2[/tex]

    Find x(6) and x(1). Then find the factor k such that x(6)=k*x(1). Makes sense?
     
  4. Sep 4, 2006 #3

    EP

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    Use your kinematic equation for distance.
     
  5. Sep 4, 2006 #4
    ok im going to say what i know

    v.=0
    t=6s
    x.=0
    a=unknown
    but what would final x be?
     
  6. Sep 4, 2006 #5

    quasar987

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    final x is given by the equation of kinematics.. this is what it says: "give me an initial position, initial velocity and a time and I'll give you a final x"
     
  7. Sep 4, 2006 #6
    but how can i use the equation when i dont know a, or final x and both are in the equation?
     
  8. Sep 4, 2006 #7

    EP

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    a is constant right? So its just a proportion. The answer will have accelerration in it.
     
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