The car will halt at a minimum distance of 98.57 meters on a rainy day having a friction coefficient of 0.109.
Calculation of the minimum distance-
Provided that :
= 52 x 0.278 m/s = 14.45 m/s
the regular force exerted on the car,
[tex]F_{N} = mg[/tex]
Along X-direction, the force is
[tex]F_{x} = ma_{x}[/tex]
Friction acts in the opposite way along the x-axis
[tex]-f_{friction} = ma_{x}[/tex]
⇒-μ[tex]F_{N} = ma_{x}[/tex]
⇒-μmg = [tex]ma_{x}[/tex]
⇒[tex]a_{x} =[/tex] μg
Utilizing the motion equation-
v² = u²+ 2 .a. s
The final speed, v=0 m/s
⇒0² = (14.45)² - 2 μg .s
⇒2 * 0.109 *9.8 *s = (14.45)² = 208.8
⇒s = 208.8 / 2.14 = 97.57 m
It is concluded that the car will halt at a minimum distance of 98.57 meters.
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