John K Clark wrote:
On Fri, 10 Jul 1998 John Heritage <firstname.lastname@example.org>
> >1 G in acceleration is 32 feet per second, per second, right?
> >That means if I'm accelerating in my race car (which I plan to do
> >tonight at the local drag strip), with slicks.. if I cover 32 feet
> >in < 1 second, I've pushed for more than 1 G "off the line", and if
> >I cover 96 feet in < 2 seconds, I've accelerated at more than 1 G
> >for the first 100 feet, basically, right?
> Distance = Acceleration times Time squared so if you maintain 1 g after one
> second you've gone 32 feet , after 2 sec 128 feet, after 3 sec 288 feet.
Err, as I recall, distance equals HALF of acceleration times the square of the time. s = .5 a t^2
That's 16 feet in the first second, 64 after two seconds, etc. I think the original question related to the coefficient of friction of the "slicks". If the upper bound on coefficient of friction were 1, then you could not accelerate at >1g on a flat surface on earth, or drive from a standing start up a > 1/1 (45 degree) slope. I was asleep that day in class, so I don't remenber. However, my kid's tech class did a hill-climber design contest. I'm pretty sure the best climbers were doing just less than a 2/1 (60 degree) slope. This was with styrofoam wheels on a sandpaper slope.