A car travels at 64 mi/h when the brakes are suddenly applied. Consider how the tires of a moving car come in contact with the road. When the car goes into a skid (with wheels locked up), the rubber of the tire is moving with respect to the road; otherwise, when the tires roll, normally the point where the tire contacts the road is stationary. Assume the coefficients of friction between the tires and the road are ìK = 0.80 and ìS = 0.90.
(a) Calculate the distance required to bring the car to a full stop when the car is skidding.
(b) Calculate the distance required to bring the car to a full stop when the wheels are not locked up.
c) How much farther does the car go if the wheels lock into a skidding stop? Give your answer as a distance in meters and as a percent of the nonskid stopping distance
(d) Can antilock brakes make a big difference in emergency stops? Explain.
To calculate the distance required to bring the car to a full stop in each scenario, we can use the formula for the distance traveled during constant acceleration:
1. When the car is skidding (wheels locked):
In this case, the car is skidding, and the coefficient of kinetic friction (μK) applies. The equation we can use is:
v^2 = u^2 + 2as
where:
- v is the final velocity (0 in this case)
- u is the initial velocity (64 mi/h = 94.66 ft/s)
- a is the acceleration (μK * g)
- s is the distance traveled
Let's convert the initial velocity to feet per second:
94.66 ft/s
The acceleration is given by the equation:
a = μK * g
Substituting the values:
a = 0.80 * 32.174 ft/s^2
Now we can solve for the distance traveled:
0 = (94.66 ft/s)^2 + 2 * (0.80 * 32.174 ft/s^2) * s
Simplifying the equation gives:
0 = 8977.3156 ft^2/s^2 + 51.518656 ft/s^2 * s
Rearranging the equation to isolate 's', we get:
s = -8977.3156 ft^2/s^2 / (51.518656 ft/s^2) = -174.3 ft
The negative sign indicates that the distance is in the opposite direction of motion. However, we are only interested in the magnitude of the distance, so we take the absolute value:
The distance required to bring the car to a full stop when skidding is approximately 174.3 ft.
2. When the car is not skidding (wheels rolling):
In this case, the car is not skidding, and the coefficient of static friction (μS) applies. We can use the same formula as before, but with the different coefficient of friction:
a = μS * g
a = 0.90 * 32.174 ft/s^2 = 28.9566 ft/s^2
0 = (94.66 ft/s)^2 + 2 * (28.9566 ft/s^2) * s
Simplifying the equation gives:
0 = 8977.3156 ft^2/s^2 + 57.9132 ft/s^2 * s
Rearranging the equation to isolate 's', we get:
s = -8977.3156 ft^2/s^2 / (57.9132 ft/s^2) = -155.0 ft
Taking the absolute value:
The distance required to bring the car to a full stop when the wheels are not locked up is approximately 155.0 ft.
c) The difference in distances:
To find how much farther the car goes if the wheels lock into a skidding stop, we subtract the stopping distance with locked wheels from the stopping distance without locked wheels.
155.0 ft - 174.3 ft = -19.3 ft
Again, we take the absolute value:
The car goes approximately 19.3 ft farther if the wheels lock into a skidding stop.
Now, let's convert the distances to meters:
1 ft = 0.3048 m
-174.3 ft * 0.3048 m/ft = -53.10 m
-155.0 ft * 0.3048 m/ft = -47.24 m
-19.3 ft * 0.3048 m/ft = -5.89 m
The car goes approximately 53.10 m farther if the wheels lock into a skidding stop.
To find the percent difference, we divide the difference in distance by the stopping distance without locked wheels and multiply by 100:
(-5.89 m / 47.24 m) * 100 = -12.46%
The car goes approximately 12.46% farther if the wheels lock into a skidding stop.
d) Antilock brakes:
Antilock brakes can make a big difference in emergency stops. They help prevent the wheels from locking up, allowing the tires to maintain better traction and control. By modulating the brake pressure rapidly, they allow the driver to steer and maintain stability while coming to a stop. This can significantly reduce the braking distance and improve the overall control of the vehicle during emergency situations.