A 1100 kg car moves along a horizontal road at speed v0 = 21.7 m/s. The road is wet, so the static friction coefficient between the tires and the road is only µs = 0.144 and the kinetic friction coefficient is even lower, µk = 0.1008. The acceleration of gravity is 9.8 m/s 2 . What is the shortest possible stopping distance for the car under such conditions? Use g = 9.8 m/s 2 and neglect the reaction time of the driver. Answer in units of m.

The car is moving at a constant velocity, so it's not accelerating, on the other side we have a friction force, this force is decelerating the car, so let's analyze it:

[tex]F_f=\µ*F_N\\F_N=m*g\\F_f=\µ*m*g\\F_f=m*a_{max}\\so:\\m*a_{max}=\µ*m*g\\a_{max}=\µ*g[/tex]

in this case, we need to use the kinetic friction coefficient because the car is moving:

[tex]a_{max}=0.1008*9.8m/s^2\\a_{max}=0.99m/s^2[/tex]

The car is decelerating at [tex]0.99m/s^2[/tex]

to obtain the distance we can use:

[tex]V_f^2=V_o^2+2a*x\\0=(21.7m/s)^2-2*(0.99m/s^2)*x\\x=238m[/tex]