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
Vf^2=Vi^2+2ad
but a= force/mass= mu*mg/m= mu*g
solve for distance d.
To find the minimum stopping distance of a car, we need to consider the force of friction acting on the car. The force of friction opposes the motion of the car and eventually brings it to a stop.
The force of friction can be calculated using the equation:
Frictional force (Ff) = coefficient of friction (µ) * normal force (N)
(a) When the coefficient of static friction between the road and tires on a rainy day is 0.100, the normal force is equal to the weight of the car. The weight of the car can be calculated using the equation:
Weight (W) = mass (m) * acceleration due to gravity (g)
The acceleration due to gravity is approximately 9.8 m/s^2.
Therefore, the normal force (N) = W = m * g.
Once we have the normal force, we can calculate the frictional force using the equation mentioned earlier.
Frictional force (Ff) = 0.100 * N
To bring the car to a stop, the frictional force must provide a deceleration equal to the car's initial velocity squared divided by twice the stopping distance. In other words:
Ff = (0.5 * m * v^2) / d
Where:
m is the mass of the car
v is the initial velocity of the car
d is the stopping distance
Now, we can equate the equation for frictional force to the equation for deceleration:
0.100 * N = (0.5 * m * v^2) / d
Rearranging the equation, we can solve for the stopping distance (d):
d = (0.5 * m * v^2) / (0.100 * N)
Substituting the given values:
v = 45.0 mi/h (convert to m/s)
µ = 0.100
m is not given (so we cannot calculate the exact distance)
To convert the velocity from miles per hour to meters per second, we can use the conversion factor:
1 mile/h = 0.44704 meters/second
Therefore, v = 45.0 mi/h * 0.44704 m/s = 20.1168 m/s
Now, let's solve for the minimum stopping distance in terms of "m" (as mass is not given):
d = (0.5 * m * (20.1168 m/s)^2) / (0.100 * N)
(b) Similarly, we can solve for the stopping distance when the surface is dry and µs = 0.602. Using the same approach and equations mentioned above, we can substitute the given values into the equation to find the stopping distance.
S=u^2/2µs(g)
S=(20.1168m/s)^2/2*.602*9.8
S=34.298m
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