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Calculating v from v0, frictional coefficients, and angle

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  1. Feb 10, 2013 #1
    You testify as an expert witness in a case involving an accident in which car A slid into the rear of car B, which was stopped at a red light along a road headed down a hill. You find that the slope of the hill is θ = 12.0°, that the cars were separated by distance d = 23.0 m when the driver of car A put the car into a slide (it lacked any automatic anti-brake-lock system), and that the speed of car A at the onset of braking was v0 = 20.0 m/s. With what speed did car A hit car B if the coefficient of kinetic friction was (a) 0.560 (dry road surface) and (b) 0.190 (road surface covered with wet leaves)?


    equations:
    Fn=mg
    Fk=μkFn
    x-xo=vot-(1/2)at^2


    attempt at solution:
    Fn=mgcos12
    =(9.8)(cos12)(m)
    =9.59m

    fk=μkFn
    fk-mgsin12=ma
    μkgcos12-gsin12=a
    (.560)(9.8)(cos12)-(9.8)(sin12)=a
    a=3.33

    x-xo=vot-(1/2)at^2
    23=20t-(1/2)(3.33)t^2
    0=-1.665t^2+20t-23

    quadratic formula gave -1.288, -10.72

    v=vo+at
    v=20+(-3.33)(1.05)
    v=16.5035

    obviously can't have -t so I just plugged in +1.05 into that last equation, but I'm just trying to figure out what I did wrong because this was the wrong answer. Thank you!
     
  2. jcsd
  3. Feb 10, 2013 #2

    TSny

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    Which direction are you taking to be the positive direction along the slope? Do both of your signs for the initial velocity and acceleration (in the x equation) agree with your choice of positive direction?

    [EDIT: Never mind, I saw a = 3.33 (positive) but you took care of the sign by inserting a negative in the x formula. Sorry]
     
  4. Feb 10, 2013 #3

    TSny

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    Are you sure you get negative values for t when you solve the quadratic equation? I get positive values.
     
  5. Feb 10, 2013 #4
    Oh--you're right! Recalculated with an online quad formula calculator and it worked. Thank you for your help!
     
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