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Homework Help: Motion at constant acceleration problem

  1. Aug 21, 2010 #1
    In coming to a stop, a car leaves skid marks 92 m long on the highway. Assuming a constant deceleration of 7.00 m/s^2, estimate the speed of the car just before breaking.

    I think my knowns are the initial position (0 m), final position (92m), acceleration (7.00 m/s^2) and final velocity (0 m/s). But I don't have the time it took to slow down and idk, none of the equations I have for motion at constant acceleration really look like they would help me solve the problem. I apologize for this silly question because I just learned about motion at constant acceleration a few days ago and so I'm not really good at this at all :/

    Thank you so much for helping!
     
  2. jcsd
  3. Aug 21, 2010 #2

    Redbelly98

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    Other than the initial position, three pieces of information are always sufficient to solve a (one-dimensional) constant acceleration problem. Since you have three pieces of information -- final position, acceleration, and final velocity -- this is solvable.

    Try having another look at those equations. You want to find one that contains final position, acceleration, and final velocity in it.

    p.s. Remember to watch the +/- signs. Are you sure the acceleration is what you said it is? :wink:
     
  4. Aug 21, 2010 #3

    ehild

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    What equations you have for constant acceleration? You have to use them, just think of how.
     
  5. Aug 21, 2010 #4
    Would the acceleration be considered negative? If the car is slowing down, its acceleration is not always necessarily negative, but would this be a situation where it is?

    The equations I have:

    v= vi + at
    x= xi + vi*t + (1/2)(a)(t)^2
    v^2 = vi^2 + 2a(x-xi)
     
    Last edited: Aug 21, 2010
  6. Aug 21, 2010 #5

    Redbelly98

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    The idea is to pick a direction to be positive, and be consistent. If you pick the direction the car is traveling to be positive, then yes the acceleration is negative.

    Good. Which one of those contains displacement, final velocity, and acceleration?
     
  7. Aug 21, 2010 #6
    Would it be the last one? I think that taking the acceleration to be negative would help, I tried this one before but got some outlandish number :P
     
  8. Aug 21, 2010 #7

    OmCheeto

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    That outlandish number wouldn't have been equivalent to 2881 miles per hour would it?
     
  9. Aug 21, 2010 #8

    Redbelly98

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    Yes.
     
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