Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: 0-60mph back down to 0 = X feet

  1. Nov 8, 2007 #1
    Acceleration.... HELP! ASAP

    I am going to court tomorrow for a trafic citation and need to figure out the following

    0-60mph back down to 0 = X feet

    information i have:

    0-60 = 6.7 sec

    60-0 = 112 ft
  2. jcsd
  3. Nov 8, 2007 #2
    As I see it, your calculation depends on the coefficient of friction for your tires, which varies widely for peeling out and skidding vs "normal" driving. I am not a lawyer, but I believe such a calculation is not easily proven. Perhaps if you take the limiting case of the static(?) coefficient of friction for the maximum acceleration allowed, at least you might argue against reckless driving (peeling out or skidding). Also, in that case, the equality of maximum acceleration and deceleration would yield tacc=tdec for elapsed time and Xacc=Xdec for distance covered. If you have skid marks, you must know the kinetic(?) coefficient of friction for your tires, which can have substantial wiggle room, although the police may have standard calculations for determining infractions.
    Last edited: Nov 9, 2007
  4. Nov 9, 2007 #3
    holy sH#T, I should have mentioned that am i am only 17 and struggling to pass my alg trig class. I think i under stood two words!

    Also the take off was normal, no burn out/ wheel spin
    Last edited: Nov 9, 2007
  5. Nov 9, 2007 #4
    Assuming constant and equal acceleration from zero to sixty and equal deceleration from sixty to zero

    a=2X/t2=86.5 ft/s2

    Whether this acceleration/deceleration is reckless, ask the experts - the police! There is no definite "acceleration limit" that I know of.
  6. Nov 9, 2007 #5
    this is what i got from a friend
    From Road & Track Cayman test (Jan. 07) the base Caymn goes 60mph to 0 in 110ft.

    60 Mph = 88 ft/sec. figure 1/2 sec for reaction time = 44 ft. so we are up to 154 ft.

    Now for the acceleration. Assume that it is constant so velocity v(t) = (88.0/6.7) * t

    So the distance is d(t) = ((88.0/6.7) * t * t )/2 at t= 6.7 sec. d = 44.0 * 6.7 = 294 ft.

    Another way to think of this is that the average velocity is half of the max velocity = 88.0 / 2 = 44 ft/sec. then the distance is 44.0 * 6.7 – same answer as the integration of the velocity equation.

    So we get 294 + 154 = 448 ft. total
  7. Nov 9, 2007 #6
    i need to figure out away to make my distance>500ft
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook