Re: Repeal of the National Speed Limit Law
Robert T. Weverka (weverka@spot.Colorado.EDU) wrote:
: firstname.lastname@example.org (Lloyd R. Parker) writes:
: >Sorry, but an object in motion possesses an amount of KE equal to 1/2
: >times its mass times its velocity squared. Basic physics. To take the
: >object from 10 mph to 0 mph requires removing a certain amount of KE
: >equal to 1/2 times the mass times 100 (1/2m10^2 - 1/2m0^2). To take the
: >object from 20 mph to 10 mph requires removing KE equal to 1/2 times the
: >mass times 300 (1/2m20^2 - 1/2m10^2).
: So what. The force you have available is about 1 g. Limited by tire adhesion.
: At 1 g deceleration, you go from 20 mph to 10 mph in the same time it takes
: 1 g deceleration to take you from 10 mph to 0 mph.
: >The amount of force you have to use is not a constant -- it will depend
: >upon how much KE you have to remove.
: Force = time derivative of momentum = m dV/dt
: Constant force = constant deceleration.
s = vt -1/2at^2 (s=distance) since a=v/t, t=v/a
s = v(v/a) -1/2 a(v/a)^2
s = v^2/a -1/2(v^2/a)
s = 1/2(v^2/a)
So stopping distance is proportional to the square of the velocity, not
the velocity itself. And distance is what keeps you out of a collision
(or gets you into one).