A car is traveling at 50.0 km/h on a flat highway. If the coefficient of friction between road and tires on a rainy day is 0.100, what is the minimum distance in which the car will stop?

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To determine this, apply Newton&#039;s Second Law in the Y direction&amp;nbsp;FN=mg. To ascertain that, set up Newton&#039;s Second Law in the x direction&amp;nbsp;ax=−μg. Remember that&amp;nbsp;Fiction=μFN. Due to its opposition to the direction of motion, the acceleration will be negative.∑Fy=mayFN−Fg=0FN=Fg=mg∑Fx=max−Fiction=max−μFN=max−μmg=maxax=−μgPut the speed into meters per second first. then apply&amp;nbsp;vf2=vi2+2ad&amp;nbsp;the stopping distance to be determined. Use the acceleration and starting speed from the previous section.&amp;nbsp;(ax=−μg)&amp;nbsp;from above.50.0kmh⋅1000&amp;nbsp;m1&amp;nbsp;km⋅1&amp;nbsp;h3600&amp;nbsp;s=13.9msvf2=vi2+2advf2=vi2−2μgd02=13.92−2(0.100)(9.81)dd=98.5&amp;nbsp;m

A car is traveling at 50.0 km/h on a flat

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