0-60mph back down to 0 = X feet

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Homework Help Overview

The discussion revolves around calculating the distance covered when accelerating from 0 to 60 mph and then decelerating back to 0 mph. The original poster provides specific timing and distance data related to their situation, which is tied to a traffic citation.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants explore the role of the coefficient of friction in calculations, question the validity of the assumptions made, and discuss the implications of acceleration and deceleration rates. Some participants offer calculations based on provided data, while others express confusion about the concepts involved.

Discussion Status

Several participants have contributed calculations and insights, with some suggesting that the assumptions about constant acceleration and deceleration may not hold in all scenarios. There is an ongoing exploration of how to approach the problem, but no consensus has been reached on a definitive method or solution.

Contextual Notes

The original poster mentions their age and struggles with algebra and trigonometry, indicating a potential barrier to understanding the mathematical concepts discussed. Additionally, the context of a traffic citation adds urgency to the calculations being explored.

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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
 
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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.
 
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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
 
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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.
 
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
 
i need to figure out away to make my distance>500ft
 

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