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Stopping distance given Vinitial and distance

  1. Feb 27, 2010 #1
    1. The problem statement, all variables and given/known data
    It takes a minimum distance of 41.14 m to stop a car moving at 11.0 m/s by applying the brakes (without locking the wheels). Assume that the same frictional forces apply and find the minimum stopping distance when the car is moving at 25.0 m/s.


    2. Relevant equations

    I'm using:
    d=1/2(Vf + Vi) * T

    and:

    v^2 = u^2 + 2as

    3. The attempt at a solution

    So here we go, I'm going to calculate the time to stop using the first formula:
    41.14 = 1/2(0+11) * T
    T = 7.48

    Cool, makes sense, alright. Now I'm going to use that number to get the acceleration:

    0^2 = 11^2 + (2 * a * 7.48)
    a = -8.088
    Makes plenty of logical sense, am I right? Sweet,
    Now I'm taking the acceleration and figuring out the stopping time of the faster moving vehicle like so

    0^2 = 25^2+(2*-8.088*t)
    t=38.64
    This is weird, it takes 38 seconds for the car to stop, oh well let's see the distance

    d = 1/2(Vf + Vi) * t
    d = 1/2(0+25) * 38.64
    d = 483 meters

    Yeah no way....plug it in as the answer and no dice.

    What am I doing wrong? Am I completely off here? Help!
     
  2. jcsd
  3. Feb 27, 2010 #2

    rl.bhat

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    Homework Helper

    0^2 = 11^2 + (2 * a * 7.48)
    a = -8.088

    This step is wrong.
    It should be
    v^2 = u^2 -2*a*s, and s = 41.14 m
     
  4. Feb 27, 2010 #3

    mgb_phys

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    Science Advisor
    Homework Helper

    You have mixed up the equations near the end
    it's v^2 = u^2 + 2 a S , you seem to have done '2 a T'

    As a rough estimate, the ke of the car is proprtional to v^2 so double the speed is 4x the energy = 4x the stopping distance
     
  5. Feb 27, 2010 #4
    I'm an idiot! Thanks for the help!
     
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