Deceleration rate with studded snow tires

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Discussion Overview

The discussion revolves around the stopping distances of a car equipped with studded snow tires on packed powder at different speeds, specifically comparing stopping distances at 25 mph and 20 mph. Participants explore the implications of using a deceleration constant derived from initial conditions and consider alternative models for calculating stopping distances.

Discussion Character

  • Exploratory
  • Technical explanation
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • One participant calculates a deceleration constant based on stopping distance from 25 mph and proposes using it to find stopping distance at 20 mph.
  • Another participant suggests that energy dissipated during braking is proportional to the distance covered, presenting an alternative model.
  • A participant questions the validity of the initial calculation and suggests that the average velocity should be considered in the context of uniform acceleration.
  • Concerns are raised about whether sliding friction produces a constant deceleration rate, with differing views on the nature of acceleration in this context.
  • Discussion includes the relationship between kinetic energy and speed, with participants exploring how energy decreases with distance during braking.
  • Some participants express agreement on the ratios of stopping distances at different speeds, while others clarify the calculations involved.

Areas of Agreement / Disagreement

Participants express differing views on the validity of the initial calculations and the assumptions made regarding deceleration. There is no consensus on the best approach to calculate stopping distances, and multiple competing models are presented.

Contextual Notes

Participants acknowledge limitations in their calculations, such as the need for more information about the car's weight and the assumptions regarding constant deceleration. The discussion also highlights the complexity of relating kinetic energy to stopping distances.

Heidi Henkel
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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?
 
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Or you make a different assumption: the energy dissipated is proportional to the distance covered with the brakes on.
 
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...
 
Heidi Henkel said:
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?
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:
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.
 
Heidi Henkel said:
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.
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.)
 
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?
 
Heidi Henkel said:
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?
Linearly with respect to what other quantity?
 
distance
 
  • #10
Heidi Henkel said:
distance
Right.

So ##\displaystyle \ \frac{v_1^2}{v_2^2}=\frac{d_1}{d_2} \ ##
 
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  • #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
 
  • #12
ok this makes sense, the equal ratios
 
  • #13
Heidi Henkel said:
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
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.
 
  • #14
Heidi Henkel said:
ok this makes sense, the equal ratios
Good !

By the way; Welcome to PF !
 

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