A car is traveling at 66.9 mi/h 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 0.13, what is the minimum distance in which the car will stop?
(1 mi = 1.609)
Answer in units of m
To find the minimum stopping distance of the car, we need to consider the forces acting on the car.
First, let's convert the car's speed from miles per hour to meters per second:
66.9 mi/h * 1.609 km/mi * 1000 m/km * 1 h/3600 s ≈ 29.95 m/s
The only force that opposes the car's motion is the force of friction. The formula for the force of friction is given by:
Frictional force = coefficient of friction * normal force
The normal force can be calculated using the weight of the car, which is given by:
Weight = mass * acceleration due to gravity
Now, we need to find the mass of the car. To do this, let's assume the car weighs 1 ton, which is approximately equal to 1000 kg.
So, mass = 1000 kg
Substituting the values into the equation, we have:
Weight = 1000 kg * 9.8 m/s^2 = 9800 N
Now, we can calculate the frictional force using the formula:
Frictional force = 0.13 * 9800 N = 1274 N
The frictional force is what slows down the car. We can use Newton's second law to determine the deceleration of the car:
Force = mass * acceleration
1274 N = 1000 kg * acceleration
Rearranging the equation and solving for acceleration, we have:
Acceleration = 1274 N / 1000 kg ≈ 1.274 m/s^2
Now, we can use the kinematic equation to find the minimum stopping distance:
v^2 = u^2 + 2as
Where:
v = final velocity (0 m/s, as the car stops)
u = initial velocity (29.95 m/s)
a = acceleration (-1.274 m/s^2, as it opposes the car's motion)
s = distance
Rearranging the equation to solve for s, we have:
s = (v^2 - u^2) / 2a
Substituting the values, we get:
s = (0 - (29.95)^2) / (2 * (-1.274)) ≈ 557.87 m
Therefore, the minimum distance in which the car will stop is approximately 557.87 meters.