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Newtons 2nd law problem - how to interpret the break force

  1. Feb 16, 2012 #1
    1. The problem statement, all variables and given/known data

    A car travels on a horizontal street, straightforward. In some moment of time (t = 0), having a velocity of 20 m/s, the break is hit. It is required to find the time and traveled distance before the car is still standing. The friction coefficient between the wheels and the street is 0.6.

    2. Relevant equations

    Newtons 2nd law

    3. The attempt at a solution

    This is a highschool problem, that means it should be simple enough, but I'm puzzled by how to interpret the break force.
    My attempt was: if we assume that the car is moving along the positive x-axis, in the moment when the break is hit, a negative force is applied in addition to the frictional resistance, but the car still moves in the positive direction, so there must be a force which "pushes" it forward. As a result, I get t = 3.33s and s = 120m, but this is wrong.

    PS: I tried even differential calculus, and would also like to ask, how one handles
    [itex]\int\frac{1}{x}dx[/itex] in the bounds starting from 0 to some number t?
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Feb 16, 2012 #2

    tiny-tim

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    welcome to pf!

    hi scienalc! welcome to pf! :smile:

    (here's some help with your english :wink:
    A car travels on a horizontal street, straight forward. At time t = 0, at a velocity of 20 m/s, the brake is hit. It is required to find the time and the distance traveled before the car stops.)

    no, the only horizontal external force is the friction

    (the force between the brake and the wheels is an internal force, and should be ignored)
    no, that is nonsense, you must stop thinking like that :redface:

    newton's first law … a body keeps moving if there is no force … it does not need "pushing"
    how did you get that? :confused:

    start by finding the acceleration

    what is it? :smile:
     
  4. Feb 16, 2012 #3
    thank you
    actually, I was assuming that what you said, and took only the frictional force into account: ma = -(mi)mg,
    then dv = -(mi)gdt,
    and from there (first integral v1=20 m/s -> v2=0 and second t1=0 -> t2=t)
    I got v(t) = (mi)gt (1)

    then the required time i calculated as: t = v1/((mi)g) = 3.33s
    after that in the equation (1): dx = (mi)gtdt
    and got x = 0.5*v1^2/((mi)g) = 66.67m (sry don't know why i wrote 120m :D )

    PS: would that mean that the car would need the same amount of time to stop, if the break wasn't hit (the car just left to the friction force to eventually stop it)?
     
    Last edited: Feb 16, 2012
  5. Feb 16, 2012 #4

    tiny-tim

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    hi scienalc! :smile:

    (have a mu: µ :wink:)
    the 3.33 looks ok

    i don't see how you get 66.67 from your formula :confused:
    no, they're the same thing

    the brake causes the friction force that decelerates the car

    (in exam questions like this, it is always assumed that there is no friction in the bearings, and that the car will continue infinitely far at the same speed if it isn't braked! :biggrin:)
     
  6. Feb 16, 2012 #5
    thanks a lot again for the quick answer (and the µ of course) :D

    it's ok, i found my mistake, the correct value for distance should be 33.3m (you get that from the equation x(t) = (µ*g*t^2)/2 which represents traveled distance in dependence of time, and use that t = 3.33s)

    funny thing, this is what I first did, but then for some reason I bugged myself with that breaking force and all went down and become confusing
     
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