Vo = 58.2km/h = 58200m/3600s. = 16.2 m/s
a = u*g = 0.155 * -9.8 = -1.52 m/s^2.
V^2 = Vo^2 + 2a*d = 0
d = -(Vo^2)/-2a = -(16.2^2)/-3.04 = 86.3
m.
A car is traveling at 58.2 km/h on a flat highway.
The acceleration of gravity is 9.8 m/s2 .
If the coefficient of friction between the road and the tires on a rainy day is 0.155, what is the minimum distance in which the car will stop?
1 answer
1 answer