A car is traveling at 60.6 mph on a horizontal highway. The acceleration of gravity is 9.8 m/s2. If the coefficient of friction between road and tires on a rainy day is 0.056, what is the minimum distance in which the car will stop? Answer in units of m. m = 60-kg.
To find the minimum stopping distance, we need to calculate the deceleration of the car due to friction.
First, let's convert the car's speed from mph to m/s:
1 mph = 0.44704 m/s
60.6 mph = 60.6 * 0.44704 m/s = 27.108024 m/s
The force of friction opposing the motion of the car is given by:
Frictional force = coefficient of friction * normal force
The normal force can be calculated as:
Normal force = mass * acceleration due to gravity
Normal force = 60 kg * 9.8 m/s^2 = 588 N
Now, the frictional force can be calculated as:
Frictional force = 0.056 * 588 N = 32.928 N
Using Newton's second law (F = ma), we can find the deceleration of the car:
Deceleration = Frictional force / mass
Deceleration = 32.928 N / 60 kg = 0.5488 m/s^2
The minimum stopping distance can be found using the equation:
Stopping distance = (Initial velocity^2) / (2 * deceleration)
Plugging in the values:
Stopping distance = (27.108^2) / (2 * 0.5488) = 428.9472 / 1.0976 = 390.418 m
Therefore, the minimum distance in which the car will stop on a rainy day is approximately 390.418 meters.
To find the minimum stopping distance, we need to consider the forces acting on the car. The main force that opposes the car's motion is the friction force between the tires and the road. This friction force can be calculated using the formula:
Friction Force = Coefficient of Friction × Normal Force
The normal force is equal to the weight of the car, which can be calculated using the formula:
Weight = mass × acceleration due to gravity
Given that the car's mass is 60 kg and the acceleration due to gravity is 9.8 m/s^2, we can calculate the weight:
Weight = 60 kg × 9.8 m/s^2 = 588 N
Now, we can calculate the friction force:
Friction Force = 0.056 × 588 N = 32.928 N
The friction force will act in the opposite direction of the car's motion, providing a deceleration. Using Newton's second law (F = m × a), we can find the deceleration:
Deceleration = Friction Force / mass = 32.928 N / 60 kg = 0.548 m/s^2
Next, we can use the following equation of motion:
v^2 = u^2 + 2as
where:
v = final velocity (0 m/s since the car stops)
u = initial velocity (60.6 mph converted to m/s)
a = deceleration
s = stopping distance (what we need to find)
The initial velocity in m/s is:
u = 60.6 mph × 0.44704 m/s = 27.095424 m/s
Substituting the values into the equation of motion, we get:
0^2 = (27.095424 m/s)^2 + 2 × 0.548 m/s^2 × s
Simplifying the equation:
0 = 733.70372399616 m^2/s^2 + 1.096 m/s^2 × s
Rearranging the equation to solve for s:
s = - (733.70372399616 m^2/s^2) / (1.096 m/s^2)
s ≈ -669 m^2/s^2 / m/s^2
s ≈ -669 m
Since distance cannot be negative, the result of the calculation is extraneous. Therefore, we ignore the negative sign, and the minimum stopping distance of the car is approximately 669 meters.
ΔKE= W(fr)
ΔKE= m•v²/2
W(fr) = μ•m•g•s
m•v²/2=μ•m•g•s
s= v²/2 μ•g