A person who is properly constrained by an over-the-shoulder seat belt has a good chance of surviving a car collision of the deceleration does not exceed about 30 "g's" (1.0 g = 9.8 m/s2). Assuming uniform deceleration of this value, calculate the distance over which the front end of the car must be designed to collapse if a crash brings the car to a rest from 100 km/h.
2006-09-06 02:48:56 · 1 answers · asked by Anonymous in Science & Mathematics ➔ Physics
okay, i got that the compaction distance is roughly 1.3122532963804902 meters. i got that by doing the following...
-30gs (1g=9.8) = -294 max acceleration
100 KPH= 100000 MpH = 27.7778 MpS (this is your initial velocity)
0 = your final velecity.
now i looked up the good old one dimensional motion equations online (you can find the cool site i found in my sources) and found the good piece of information
V^2 = U^2 +2AS where V is final velocity, U is initial velocity, A is the acceleration, and S is the distance travelled during that time. you have all but S at this point, so solving for S gives you your answer. in fact, the page that i link to even does the calculation for you.
so there's your answer. next time do your own homework.
2006-09-06 04:17:30 · answer #1 · answered by promethius9594 6 · 0⤊ 0⤋