Ok... i have a portfolio task thingy from some kind of IB hand book or watever and i have to figure out an equation that models the data.. i've figured out the linear equation for thinking distance.... now im stuck on braking distance....
the data is:
speed braking distance
32_______6,
48_______14,
64_______ 24,
80_______38,
96_______55,
112 _______75,
2006-11-24 05:13:47 · 4 answers · asked by scoot 2 in Science & Mathematics ➔ Mathematics
....this is not a physics problem.... there is no accelaration these are indivudual speeds with individual stop times
2006-11-24 07:56:00 · update #1
Breaking distance is related to speed ^2. From the examples given, breaking distance = speed^2/168. For example, at a speed of 32, breaking distance = 32^2/168 = 6.1 and at 112, bd = 112^2/168 = 74.7.
2006-11-24 07:08:55 · answer #1 · answered by JJ 7 · 0⤊ 0⤋
I don't know what the units are for the data but the general formula that involves distance and changes in speed is:
final velocity^2 = initial velocity^2 + 2*a*d where a is the acceleration and d is the distance. You can determine the acceleration (negative in this case for slowing down) by rearranging the formula to get a = (Vf^2 - Vi^2) / 2d
2006-11-24 05:34:59 · answer #2 · answered by cheeseballer 3 · 0⤊ 0⤋
According to physics the simplest formula to approximate these data is v^2=2*s*a,
where v is your speed, s is braking distance, while ‘a’ is a parameter to suit the formula the best way; the simplest way to find a suitable ‘a’ is: find a=v^2/(2s) for each set of your 6 data, then find a as an average of 6 values of a.
to be more professional you must use regression analysis.
2006-11-24 07:29:50 · answer #3 · answered by Anonymous · 0⤊ 0⤋
Try Google.
2006-11-24 05:23:48 · answer #4 · answered by ag_iitkgp 7 · 0⤊ 0⤋