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Equation of motion

  1. May 9, 2005 #1
    a car starts from rest and acclerates down a straight track of length L= 1600m with a constant accleration. If the time it takes for the car to travel the final d= 100 m of the track (from 1500m to 1600m) is T=0.125s, then the acceleration of the car is.... the answer is 80.7 m/s^2

    this isnt hw, it's actually practice for the final i have tommorow. I havent done this question in awhile, so im not sure how to do it. but here's what i think i should do... im having trouble finding which equation to use. well i have the initial x and final x, and t. and i need to find acceleration. so i think i need to use this formula:

    [tex]X = x_0 + v_0*t + 1/2at^2[/tex]

    100 = 1500 + 1/2at^2

    x final should be 100 right? because the final 100 it needs to travel
    and solve for a? doesnt really make sense. can someone offer some help on which equation i should use?
    Last edited: May 9, 2005
  2. jcsd
  3. May 9, 2005 #2


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    You know you are starting from rest, so you can take

    [tex]x = x_0 + v_0*t + \frac{1}{2}at^2 = 0 + 0 + \frac{1}{2}at^2[/tex]

    You know two distances for x (1500, 1600), and you know the time difference between them. If t is the first time, then t + .125s is the second time. Put your times into the equation with the known distances and you will have two equations for two unknowns that you can solve for a and t.
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