v = vi - 33 t
v = 0 at stop
so
0 = 28 - 33 t
t = .848 seconds
distance = vi t -(1/2)(33) t^2
d = 28(.848) - 16.5 (.848)^2
= 23.8 - 11.9
= 11.9 meters (very long car)
Can anyone help with which formula I should use to solve this:
During a head on car crash, the car is designed to slow down as its front end crumples. If a person can survive an acceleration of -33 m/s how far must the front end crumple if the car was initially traveling at 28 m/s?
1 answer
1 answer