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Homework Help: [attempt?] car accident problem.

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  1. Mar 31, 2008 #1
    [SOLVED] [attempt?] car accident problem.

    1. The problem statement, all variables and given/known data
    http://hypertextbook.com/physics/mechanics/momentum-two-three/accident-reconstruction-1.pdf [Broken]


    2. Relevant equations
    Pix=Pfx
    Piy=Pfy
    v^2 = v0^2 + 2 ax
    Fk=MkN
    W=mg
    F=ma

    3. The attempt at a solution

    01.
    Fk = MkN
    100 N = Mk (130 N)
    0.769 = Mk

    car tires = 0.769 --- but then, i'm not sure of the units.


    0.769 (0.7) = 0.538

    truck tires = 0.538 --- again with the units.

    02.
    W = 13600 N
    W = mg
    13600 = m (9.8 m/s^2)
    1387.755 kg = m

    Fk = MkN
    Fk = 0.769 (13600 N)
    Fk = 10458.4

    F = ma
    10458.4 = 1387.755 a
    7.536 m/s^2 = a

    v^2 = v0^2 + 2 ax
    0 = v0x^2 + 2 (-7.536) (8.2)
    123.59 = v0x^2
    11.117 m/s = speed of car


    W = 69700 N
    W = mg
    W = 7112.245

    Fk = MkN
    Fk= 0.538 (69.700)
    Fk = 37498.6

    F = ma
    37498.6 = 7112.245 a
    a = -5.272 m/s^2

    v^2 = v0^2 + 2 ax
    0 = v0^2 + 2 (-5.272) (11)
    115.984 m/s = v0^2
    10.769 m/s = speed of truck

    03.
    Pfx=Pix --- i know i wrote the other way earlier, but i'm copying my weird notes.
    (1387.755) (11.117) cos (33) + (7112.245) (10.769) cos (7) = 7112.245 v
    88958.823 = 7112.245 v
    v of truck = 12.508 m/s


    (1387.755) (11.117) sin (33) + (7112.245) (10.769) sin (7) = 1387.755 v
    17736.701 = 1387.755 v
    v of car = 12.781 m/s

    I asked my teacher and he said he vaguely remembering that the velocities were supposed to be 11?, but he and I couldn't find fault with the set up. I tried recalculating, and I came up with the same answer, which frustrates me a bit.

    04.
    As for number four, I read the facts given repeated times, but I'm still at a loss for how to approach it. I realize that I'm supposed to use parts 01.-03. though.



    Thanks. :D
    Sorry, it's a bit long.
     
    Last edited by a moderator: May 3, 2017
  2. jcsd
  3. Jun 19, 2015 #2
    The measurement of the distance between the traffic light and the collision point (d = 13.0 m) and the maximum acceleration of the car (a = 3.0 m/s^2) are not consistent with the claim of the car driver. Assuming constant acceleration, which is a pretty good assumption since the mass of the car does not change significantly as a result of the collision, we get:


    vf = Sqrt(2ad) = Sqrt[ 2(3.0m/s^2)(13.0 m) ] = 8.83 m/s.


    The percent difference between this speed and the speed of the car just before the collision calculated in question 3 is significant.


    % Difference = [ 12.781 m/s - 8.83 m/s ]/[ 12.781 m/s + 8.83 m/s ] x 200 = 37%.
     
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