Deceleration time for studded snow tires

[Heidi Henkel]

2ad=v^2
ad=.5v^2

m=1/a? mass is inversely related to acceleration, I believe that...but m=1/a, not necessarily.

 

[OmCheeto]

Energy. The kinetic energy from the moving car is dissipated by the friction force the road exerts on the tyres.

Really? I just rediscovered the "kinematics equations" about a month ago. They are a godsend for idiots like myself.

if v_0^2 = 0
then the following kinematic equation
v^2 = v_0^2 + 2 a \Delta x
reduces to
v^2 = 2 a \Delta x
which rearranged becomes
\Delta x = v^2 / 2 a

hmmmm... kinetic energy:
ke = 1/2 m v^2
f = m a --> m = f / a
therefore
ke = 1/2 (f/a) v^2 = f v^2 / 2a
and
W = \Delta ke = fd
rearranged yields
f=ke/d
and since ke_f = 0
we get
ke = ( ke/d ) v^2 / (2a)
multiplying both sides by d/ke yields, once again
d = v^2 / 2a

Ha! That works too. Thanks!

[edit: Why do I always catch my errors after pushing the post button...]

 

[OmCheeto]

Energy is not part of that equation.

That confused the heck out of me too! But I know from experience, that they don't give out Homework Helper badges to just anyone.

 

[BvU]

It's not really rocket science... (which itself is rather overrated in my opinion ) and a badge (or a PhD, for that matter) is no guarantee against stupidities.

You can check wikipedia or hyperphysics for the formulas.

Interesting reading, this 2002 pdf. I notice Table 5 and table 6 show big differerences in deceleration/acceleration.
Table 5:
If we look at the snow ones, the deceleration is around $${ (11 {\rm\ m/s}) ^2\over 2 \times (20 {\rm \ m})} \approx 3 {\rm \ m/s^2} $$ using this $$ {1\over 2} mv_{\rm end}^2 - {1\over 2} mv_0^2 = F_{\rm friction} \times d_s = - ma\times d_s$$ energy dissipation equation . That is pretty good !
The bare pavement ones are plain excellent : $\approx 12 {\rm \ m/s^2}$ ! Hurray for ABS (you did notice that in the text ?)

(question: the 20 feet skid marks in the snow - were they from a vehicle with ABS ? What tyres did it have ?)

Table 6:
Even easier to calculate acceleration: a mere $\approx 1.2 {\rm \ m/s^2}$ ! No traction control, apparently ? (the opposite of ABS).

--

I am puzzled about table 12: much more deceleration, and no explanation in the text. Makes one suspect that friction is better at higer speeds...

--