Vf^2=Vi^2+2ad
but a= force/mass= mu*mg/m= mu*g
solve for distance d.
A car is traveling at 45.0 mi/h on a horizontal highway.
(a) If the coefficient of static friction between road and tires on a rainy day is 0.100, what is the minimum distance in which the car will stop?
____ m
(b) What is the stopping distance when the surface is dry and µs = 0.602?
_____ m
3 answers
45 mi/h = 20.1168 m/s
ma= .100n
a = .100(g)
solving for time
v=vo + at
20.1168m/s=0+.98(t)
t=20.527s
solving for distance
d=do+vot+at^2/2
d= 0+0+.98(20.527s)^2/2
d= 206.469m
ma= .100n
a = .100(g)
solving for time
v=vo + at
20.1168m/s=0+.98(t)
t=20.527s
solving for distance
d=do+vot+at^2/2
d= 0+0+.98(20.527s)^2/2
d= 206.469m
S=u^2/2µs(g)
S=(20.1168m/s)^2/2*.602*9.8
S=34.298m
S=(20.1168m/s)^2/2*.602*9.8
S=34.298m
0 answers