Final answer:
- a) The minimum distance in which the car will stop on a rainy day is approximately 472.04 feet.
- b) The stopping distance when the surface is dry and the coefficient of friction is 0.597 is approximately 82.2 feet.
Step-by-step explanation:
The minimum distance in which the car will stop can be calculated using the formula:
Stopping distance = (initial velocity²) / (2 * coefficient of friction * acceleration due to gravity)
(a) To find the minimum stopping distance on a rainy day, where the coefficient of static friction is 0.105, we need to first convert the velocity from miles per hour to feet per second.
- 1 mile = 5280 feet
- 1 hour = 3600 seconds
54.0 miles/hour * 5280 feet/mile * 1 hour/3600 seconds = 79.2 feet/second (approximately)
Now, we can substitute the values into the formula:
Stopping distance = (79.2 feet/second)² / (2 * 0.105 * 32.2 feet/second^2)
Simplifying the equation gives us:
- Stopping distance = 3184.64 feet / (6.744 feet/second²)
- Stopping distance ≈ 472.04 feet
Therefore, the minimum distance in which the car will stop on a rainy day is approximately 472.04 feet.
(b) To find the stopping distance when the surface is dry and the coefficient of friction is 0.597, we can use the same formula.
First, let's convert the velocity from miles per hour to feet per second:
54.0 miles/hour * 5280 feet/mile * 1 hour/3600 seconds = 79.2 feet/second (approximately)
Now, substitute the values into the formula:
Stopping distance = (79.2 feet/second)² / (2 * 0.597 * 32.2 feet/second^2)
Simplifying the equation gives us:
- Stopping distance = 3184.64 feet / (38.7508 feet/second²)
- Stopping distance ≈ 82.2 feet
Therefore, the stopping distance when the surface is dry and the coefficient of friction is 0.597 is approximately 82.2 feet.