A car is traveling at 51.0 mi/h on a horizontal highway
If the coefficient of static friction between road and tires on a rainy day is 0.096, what is the minimum distance in which the car will stop
What is the stopping distance when the surface is dry and µs = 0.603?
By using static friction coefficient Uk, we assume that the car does NOT go into a skid. The maximum static friction amount is applied to the tires even though they are still turning. That force decelerates the car over a distance X
friction work = Uk*M g X = (1/2) M V^2
X = V^2/(2 g Uk)
V = 51 mph = 74.8 ft/s = 22.8 m/s
Solve for X, using V in m/s and g = 9.8 m/s^2, or V in ft/s and g = 32.2 ft/s^2
To find the minimum stopping distance of a car, we can use the formula:
Stopping distance = (Initial velocity ^ 2) / (2 * deceleration)
In this case, the deceleration is determined by the coefficient of static friction between the road and the tires.
For the first question, when the surface is wet with a coefficient of static friction (μs) of 0.096:
1. First, we need to convert the velocity from miles per hour to feet per second. Since there are 5280 feet in a mile and 3600 seconds in an hour, we can use the conversion factor: 1 mile/hour = 5280 feet/3600 seconds.
Therefore, the velocity is 51.0 * 5280 / 3600 = 74.8 feet/second.
2. The deceleration can be calculated using the formula:
Deceleration = coefficient of static friction * acceleration due to gravity (g = 32.2 ft/s^2).
Therefore, deceleration = 0.096 * 32.2 = 3.0912 ft/s^2.
3. Now, we can substitute the values into the stopping distance formula:
Stopping distance = (74.8^2) / (2 * 3.0912) = 1458.02 feet.
Therefore, the minimum distance in which the car will stop on a wet road is approximately 1458.02 feet.
For the second question, when the surface is dry with a coefficient of static friction (μs) of 0.603:
1. Repeat step 1 to calculate the velocity in feet per second. The velocity remains the same at 74.8 feet/second.
2. Repeat step 2 to calculate the deceleration using the new coefficient of static friction:
Deceleration = 0.603 * 32.2 = 19.4658 ft/s^2.
3. Substitute the values into the stopping distance formula:
Stopping distance = (74.8^2) / (2 * 19.4658) = 114.6499 feet.
Therefore, the stopping distance on a dry road with a coefficient of static friction (μs) of 0.603 is approximately 114.6499 feet.