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Stopping distance with nonconstant deceleration

  1. Jun 25, 2006 #1
    Another question
    It says:
    A car is travelling at 72km/hr. At a certain instant its brakes are applied to produce a nonconstant deceleration of s"(t)=-t(in m/s^2) How far does the car travel before coming to rest?

    Now theres a lot of questions like this in the book, however they use a constant deceleration which is given along with the velocity.

    but what would i use as my deceleration? I believe there probably is, but to me now it seems like there isn't enough information to answer the question. Im not asking for the solution because i know thats not what your here for, but could someone give me a hint at what I do with this nonconstant deceleration? Thanks
  2. jcsd
  3. Jun 25, 2006 #2


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    Last edited: Jun 25, 2006
  4. Jun 25, 2006 #3
    still confused???

    Sorry.. this question is driving me crazy:frown:
    I just cant seem to understand it.. ive been starring at that website, and nothings making sense to me. Do I need to figure out the time first? because by those equations thats whut its looking like... But I cant figure out how to do that either.. This questions is just confusing me soo much ... Is there anything else you could say to help me out? Thanks
  5. Jun 25, 2006 #4


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    Yes, you need to determine the time first, you can do this by finding the integral;

    [tex]v(t) = \int -t \;dt[/tex]

    What is v before the breaks are applied? What is v when the car is stopped?

    Edit: There's no need to apologise, we're all here to learn :smile:
    Last edited: Jun 25, 2006
  6. Jun 26, 2006 #5
    I still dont get it!

    Ok this could be very wrong, Im just not sure how to do it any other way.
    (im going to use $ for the integral sign)

    v(t)=$-t dt

    =-1/2 t^2

    so v(t)=-1/2 t^2 ( and ive changed km/h to m/s)
    20m/s=-1/2 t^2
    t=sq root( 20/-1/2) t=sq root(40) t= 6.324s

    Since s"(t)=-t in (m/s^2)
    does that just mean that the deceleration is -6.324m/s^2???

    t is $$-t dt so -1/3t^3
    and v(t)=-1/3t^3
    t=sq root (20/-1/3) t=sq root(60) t=7.75s

    is v=d/t d=v*t = 20m/s*7.75s=154m

    This definetly is not right... im soooo confused i have no idea whats going on here, please HELP!!!! :frown:
  7. Jun 26, 2006 #6


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