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Homework Help: SUVAT fairground ride problem

  1. Apr 13, 2012 #1
    A fairground ride ends with the car moving up a ramp at a slope of 30 degrees. Given that the deceleration of the car is 4.905 m/s, and that the car enters the bottom of the ramp at 18 m/s, calculate the minimum length of the ramp for the car to stop before it reaches the end.

    It seems if I take different methods I get two different answers, one a factor of three out from the other.

    If I try to first work out the time taken, I do as follows..

    V=u+at, so 0=18+(-4.905t), so t=3.669724771

    S=ut+1/2at^2, so s=18x3.66..+1/2*4.905*3.66..^2

    So s = 99.08256881

    BUT if I use..
    v^2 = u^2 + 2as
    I get as follows..

    0^2 = 18^2 + 2*4.905*s
    So s = 33.02752294

    Obviously one of these methods must have a flaw somewhere, but I can't for the life of me figure out where! Any help would be greatly appreciated! Thanks!
    Last edited: Apr 13, 2012
  2. jcsd
  3. Apr 13, 2012 #2


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    a is negative. Replace + by minus.

  4. Apr 13, 2012 #3
    Ah of course! Thanks so much, that really helps.
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