a typical person can tolerate an acceleration of about -49m/s2 (forward). If you are travelling at 110km/h and have a collison with a solid immovable object, over what minium distance must you stop so as to not exceed this acceleration

Answer 1

Vi = 110,000 meters / 3600 seconds

v = Vi + a t
at end, v = 0
so
0 = Vi -49 t
t = Vi/49

then
d = Vi t + (1/2) a t^2
d = Vi (Vi/49) - (1/2)(49)(VI^2/49^2)
d = (1/2)Vi^2/49