108,000m/hr * 1hr/3600s = 30 m/s
friction force= -.9 m g
= -.9 * 9.81 m = -8.83 m
F = m a
-8.83 m = m a
a = -8.83 m/s^2
v = Vi + a t
0 = 30 - 8.83 t
t = 4 seconds
8] Find the minimum stopping distance for a car moving at 108km/h if the coefficient of static friction between the tires and road is a) 0.9
1 answer