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Calculating braking distance of a car

  1. Aug 15, 2016 #1
    1. The problem statement, all variables and given/known data
    A car that weights 600kg, travelling at 300km/hr and the coefficient of static friction between the car and the road is 1.5, down force is equal to 2.6v^2 and air resistance is 1.5v^2 (v is in m/s). Calculate how many meters it would take for the car to stop with the breaks applied to there maximum value.

    2. Relevant equations
    V(final) = V(initial) + at
    Work = -1/2mv^2
    I'm not sure if there are more?

    3. The attempt at a solution
    N(normal force)= mg + 2.6v^2 when v = 83.3 (300km/hr converted to m/s)
    N= 23921.1N
    F(friction)= 1.5 * N
    = 32881.7N
    F(air resistance)= 1.5v^2 when v= 83.3
    =10408.3N
    Finding the acceleration of the forces
    a= F/m
    Total accelection = -77.3m/s^2
    How long it takes to go from 300 to 0

    0=83.3-77.3*t
    t=1.1sec
    Integration of velocity formula to get the displacement formula
    D=83.3*t-(77.2/2)*t^2 when t = 1.1
    =44.9m

    I feel like the deceleration is way to high (just under 8g) and I'm not sure how I'm to correct that as I feel like I have used formula incorrectly as doing from 300km/hr to 0 in just over a second can't be right,
    so any help I can get to point me in the right direction would be greatly appreciated.
    Thank You for your time
     
  2. jcsd
  3. Aug 15, 2016 #2

    NascentOxygen

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    Hi Blocker12. smiley_sign_welcome.gif

    Air resistance is not a fixed 10408 N. It is proportional to v2, this means as the car slows, air resistance force reduces rapidly.

    Similarly, friction has a velocity-dependent component, too.
     
  4. Aug 20, 2016 #3

    rude man

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    The more I look at this problem, the more bizarre it looks back at me.
    Why should the vehicle ever come to a stop? In order to stop, an opposing force is required even at zero velocity, which does not seem to be the case here.
    The "down" force is a mystery. Since no kinetic friction coefficient is given, the force of friction is unknowable. And anyway, it woiuldn't help by the above observation.
    Must be something I'm missing ...
     
  5. Aug 20, 2016 #4

    gneill

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    The mass of the car is given, hence its weight, and the additional downforce I suppose could be attributed to aerodynamics since it's proportional to the velocity squared. The coefficient of static friction is given, so if we assume that the car employs an efficient ABS, then...
     
  6. Aug 20, 2016 #5

    rude man

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    How does the down force (I agree it's probably due to vertical air pressure) figure in the problem? No kinetic friction coeff. given, and even if it was you need a countervailing force finite even at zero velocity to have the car come to a stop:
    m dv/dt + k v2 = 0 solves to
    v(t) = mv0/(m + kv0t) > 0 for all finite t
    where v0 = initial velocity.
    Static friction does not come into play until v = 0 but the vehicle never stops.
     
  7. Aug 20, 2016 #6

    PeroK

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    It's the coefficient of static friction that determines the maximum braking capability. Kinetic friction enters the equation if the car skids.
     
  8. Aug 20, 2016 #7

    gneill

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    The down force adds to the apparent weight, increasing the normal force, hence the maximal static friction.
     
  9. Aug 20, 2016 #8

    PeroK

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    This is a far more difficult problem than can be solved by plugging numbers into one equation. To solve this, you need a proper mathematical approach. In fact, because the forces depend on ##v##, you will need to generate and solve a differential equation. Do you know much about such equations?
     
  10. Aug 20, 2016 #9

    NascentOxygen

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    Since coeff of kinetic friction is always < static friction, for full marks perhaps the answer could take the form: stopping distance with brakes locked will be at least ...

     
  11. Aug 20, 2016 #10

    rude man

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    So far I considered only the tires against the pavement. But it seems the action of the brake is also to be considered, which makes the problem even more complex. Good luck everyone, I'm lost!
     
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