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A car is traveling at 56.9 km/h on a flat highway. the acceleration of gravity is 9.8 m/s 2 . if the coefficient of friction between the road and the tires on a rainy day is 0.109, what is the minimum distance in which the car will stop?

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[tex]Given:\\v=56.9 \frac{km}{h} \approx 15.81 \frac{m}{s} \\g=9.8 \frac{m}{s^2} \\\mu=0.109\\\\Find:\\s=?\\\\Solution:\\\\s=vt- \frac{at^2}{2} \\\\a= \frac{v}{t} \Rightarrow t= \frac{v}{a} \\\\s= \frac{v^2}{a} - \frac{v^2}{2a} =\frac{v^2}{2a} \\\\a= \frac{F}{m} \\\\F=T=\mu N\\\\N=mg\\\\F=\mu mg\\\\a= \frac{\mu mg}{m} =\mu g\\\\s= \frac{v^2}{2\mu g} \\\\\\s= \frac{(15.81 \frac{m}{s})^2}{2\cdot 0.109\cdot9.8 \frac{m}{s^2} } \approx 117m[/tex]

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