A 1200 kg car moves along a horizontal road
at speed v0 = 20.1 m/s. The road is wet,
so the static friction coefficient between the
tires and the road is only μs = 0.189 and
the kinetic friction coefficient is even lower,
μk = 0.1323.
The acceleration of gravity is 9.8 m/s2 .
What is the shortest possible stopping dis-
tance for the car under such conditions? Use
g = 9.8 m/s2 and neglect the reaction time of
the driver.
Answer in units of m.
To find the shortest possible stopping distance for the car under the given conditions, we need to consider both the static friction and kinetic friction forces acting on the car.
First, let's calculate the maximum static friction force that can act on the car to keep it moving without slipping. The formula for static friction force is given by:
Fs = μs * N
where Fs is the static friction force, μs is the static friction coefficient, and N is the normal force acting on the car. In this case, the normal force is equal to the car's weight, which is given by:
N = m * g
where m is the mass of the car. Plugging in the values we have:
N = (1200 kg) * (9.8 m/s^2) = 11760 N
Now we can calculate the maximum static friction force:
Fs = (0.189) * (11760 N) = 2222.24 N
This means that as long as the force required to stop the car is less than 2222.24 N, the car will not start skidding and the static friction force will be responsible for stopping it.
However, if the force required to stop the car exceeds the maximum static friction force, the car will start skidding and we will have to consider the lower kinetic friction coefficient.
Assuming the car stops due to kinetic friction, the friction force is given by:
Fk = μk * N
where Fk is the kinetic friction force and μk is the kinetic friction coefficient. Using the same normal force value as before:
Fk = (0.1323) * (11760 N) = 1556.688 N
To calculate the deceleration produced by the kinetic friction force, we use Newton's second law:
Fk = m * a
where a is the deceleration. Rearranging, we get:
a = Fk / m = (1556.688 N) / (1200 kg) = 1.297 m/s^2
The deceleration is negative since it is opposite in direction to the car's initial velocity.
Finally, to calculate the shortest possible stopping distance, we can use the following kinematic equation:
v^2 = v0^2 + 2 * a * d
where v is the final velocity (0 m/s), v0 is the initial velocity (20.1 m/s), a is the deceleration (-1.297 m/s^2), and d is the stopping distance we want to find.
Plugging in the values:
0 = (20.1 m/s)^2 + 2 * (-1.297 m/s^2) * d
Simplifying the equation:
0 = 404.01 m^2/s^2 - 2.594 m/s^2 * d
Rearranging to solve for d:
2.594 m/s^2 * d = 404.01 m^2/s^2
d = 404.01 m^2/s^2 / 2.594 m/s^2
d = 155.7 m
Therefore, the shortest possible stopping distance for the car under these conditions is approximately 155.7 meters.