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Stopping Distance and kinetic friction

  1. Sep 10, 2015 #1
    1. The problem statement, all variables and given/known data

    While traveling on the highway with your 1000kg car, at 115.2 km/h, where you’re ABS (automatic

    braking system) is disabled. This means braking is relying solely on the friction of your tires with the road when they stop spinning. A dear jumps into the road 50 meters in front of you. If the frictional force created by you slamming on your brakes is 4000N. What will your final stopping distance be? Will you

    hit the dear? Assume no air resistance.


    2. Relevant equations

    Kinematics?

    3. The attempt at a solution
    Sum of Forces in The X direction
    (Force O' Car) - (Force O' Friction) = -ma

    I solved for acceleration and got -5.8 m/s^2
    plugging these into the kinematics equation I got a time... 5.52 seconds
    plugging that into the Xf = Xo + Vox t + 1/2 a t^2 I got a distance which is wrong

    what gives? what am i doing wrong?
     
  2. jcsd
  3. Sep 11, 2015 #2

    RUber

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    Maybe look at the momentum of the car. Units will be N sec. What would be a reasonable way to find seconds to stop under friction force in N?
    I get something greater than 6 seconds.
     
  4. Sep 11, 2015 #3

    SteamKing

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    Staff Emeritus
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    We have no idea; you didn't provide your complete calculations.

    BTW, "you're" = "you are" and shouldn't be used to mean "your".
     
  5. Sep 11, 2015 #4

    RUber

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    By the way...what is the acceleration caused by 4000N on a 1000kg car? If the car is starting at a constant velocity, that means acceleration of the car is zero...so what is the acceleration in the system? It is not -5.8 m/sec^2.
     
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