Physics: A1100 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.

A1100 kg car moves along a horizontal road at speed - id11003507184, annamerryberry1016

Ответ:

josephbrowne9p18dit

08.01.2022

the shortest possible distance to stop is 238m

Explanation:

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:

$F_f=\µ*F_N\\F_N=m*g\\F_f=\µ*m*g\\F_f=m*a_{max}\\so:\\m*a_{max}=\µ*m*g\\a_{max}=\µ*g$

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

$a_{max}=0.1008*9.8m/s^2\\a_{max}=0.99m/s^2$

The car is decelerating at 0.99m/s^2

to obtain the distance we can use:

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

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