posted by Daniel Limon .
A person has a reasonable chance of surviving an automobile crash if the deceleration is no more than 30 "'s."
What distance is traveled if the person is brought to rest at this rate from 115 ?
a = 30gs = 30 * 9.8m/s^2 = -294 m/s^2.
V^2 = Vo^2 + 2a*d.
d = (V^2-Vo^2)/2a.
d = (0-1020.4)/-588 = 1.74 m.