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Deceleration rate with studded snow tires

  1. Dec 28, 2015 #1
    A car going 25 mph with studded tires on packed powder stops in 20 meters. What will be the stopping distance if it is going 20 mph?
    We do not know the mass of the car, but perhaps it is accurate to create an acceleration constant that will apply for other speeds?

    a=v/t and v=d/t so therefore by combining these we get a=v^2/d

    25mph is 11.176 m/s

    deceleration constant: a = (11.176m/s)^2 / 20m = 6.245 m/s^2

    OK so this is the deceleration constant for that set of tires on that car on that surface...

    then can we apply the deceleration constant to the other question? What will be the stopping distance at 20 mph?

    20mph = 8.94 m/s; stopping distance is 12.8m

    Is this valid?
     
  2. jcsd
  3. Dec 28, 2015 #2
    Or you make a different assumption: the energy dissipated is proportional to the distance covered with the brakes on.
     
  4. Dec 28, 2015 #3
    How is that a different assumption? Wouldn't those two models be compatible, except my way of doing it requires less data and is simpler?
    I am asking whether my way of calculating it is valid. Do you think it's not valid because of what you are saying? Doesn't seem to me like that would be the case...
     
  5. Dec 28, 2015 #4

    SammyS

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    No.

    v = d/t is valid if v refers to average velocity. Write that as vaverage = d/t .

    For uniform (constant) acceleration, it's also true that vaverage = (vinitial + vfinal) / 2 . With your choice of variables, this is vaverage = v/2 .


    Although it may appear that following PietKuip's suggestion is more complicated, the fact of the matter is that such an approach may be the simplest method.
     
    Last edited: Dec 28, 2015
  6. Dec 28, 2015 #5
    Sliding friction of snow tires on road does not produce a constant deceleration rate?

    If I try Pietkuip's suggestion, what would the equations be? What would the variables be? Would I need to know more information about the car? I could probably get the car's weight by looking up the specs online.
     
  7. Dec 28, 2015 #6

    SammyS

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    Yes, it produces constant acceleration. That is what I referred to.



    Kinetic Energy is proportional to the square of the speed.

    Energy dissipated is proportional to distance covered with the brakes on.

    (With this, you don't need to convert any units.)
     
  8. Dec 28, 2015 #7
    E=.5mv^2 so .5mv^2 decreases linearly? Since .5 doesn't change and the mass of the car doesn't change, V^2 decreases linearly?
     
  9. Dec 28, 2015 #8

    SammyS

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    Linearly with respect to what other quantity?
     
  10. Dec 28, 2015 #9
    distance
     
  11. Dec 28, 2015 #10

    SammyS

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    Right.

    So ##\displaystyle \ \frac{v_1^2}{v_2^2}=\frac{d_1}{d_2} \ ##
     
  12. Dec 28, 2015 #11
    OK so to decelerate from 20mph with the same car and tires and surface it takes about 13 meters and to decelerate from 15mph it takes a little more than 7 meters? Because starting at 25mph, at 13 meters to go, you are going 9m/s which is about 20 mph. Because at 13 meters to go, v^2 is 81.185
     
  13. Dec 28, 2015 #12
    ok this makes sense, the equal ratios
     
  14. Dec 28, 2015 #13

    SammyS

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    Much of that is good.

    Starting at 20 mph it takes about 13 meters to stop (actually 12.8 meters). Therefore, starting at 25 mph, at the 13 meter to go the speed is about 20 mph, since you know it takes about 13 meters to stop when starting at 20 mph.
     
  15. Dec 28, 2015 #14

    SammyS

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    Good !

    By the way; Welcome to PF !
     
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