Question
Assume that a particular crash test expert can withstand 18 times the acceleration due to gravity. Find the shortest distance over which her car can be brought to a stop from a speed of 78 km/h without the acceleration exceeding 18g.
Answers
Henry
Vo = 78km/h = 78000m/3600s. = 21.7 m/s.
V^2 = Vo^2 + 18g*d
d = (V^2-Vo^2)/18g = 0-(21.7^2)/-176.4 =
2.67 m.
V^2 = Vo^2 + 18g*d
d = (V^2-Vo^2)/18g = 0-(21.7^2)/-176.4 =
2.67 m.