a) Find the maximum rate of deceleration of a car on a dry, level road, assuming the coefficient of static friction between tires and pavement equals 1.4.
(b) Find the minimum stopping distance if the car has an initial speed of 27.5 m/s.
(c) Repeat parts (a) and (b) assuming a wet, oily surface for which μs = 0.11.
maximum rate of acceleration
minimum stopping distance
To find the maximum rate of deceleration (negative acceleration) of a car on a dry, level road, we can use the equation:
a = μs * g
Where:
a is the acceleration (deceleration) in m/s^2,
μs is the coefficient of static friction between the tires and the pavement,
g is the acceleration due to gravity (approximately 9.8 m/s^2).
a) In this case, μs = 1.4 and g = 9.8 m/s^2. Plugging these values into the equation, we get:
a = 1.4 * 9.8 = 13.72 m/s^2
This means the maximum rate of deceleration for the car on a dry, level road is 13.72 m/s^2.
b) To find the minimum stopping distance, we can use the equation:
d = (v^2) / (2a)
Where:
d is the stopping distance in meters,
v is the initial velocity in m/s,
a is the deceleration in m/s^2.
In this case, the initial velocity v = 27.5 m/s (given) and a = 13.72 m/s^2 (maximum deceleration). Plugging these values into the equation, we get:
d = (27.5^2) / (2 * 13.72) ≈ 48.96 m
Therefore, the minimum stopping distance for the car with an initial speed of 27.5 m/s is approximately 48.96 meters.
c) Now, let's repeat parts (a) and (b) assuming a wet, oily surface with a coefficient of static friction (μs) equal to 0.11.
a) Using the same equation as in part (a) but with μs = 0.11, we get:
a = 0.11 * 9.8 ≈ 1.08 m/s^2
So, the maximum rate of deceleration for the car on a wet, oily surface is approximately 1.08 m/s^2.
b) Using the same equation as in part (b) but with a = 1.08 m/s^2, we get:
d = (27.5^2) / (2 * 1.08) ≈ 362.73 m
Thus, the minimum stopping distance for the car with an initial speed of 27.5 m/s on a wet, oily surface is approximately 362.73 meters.