Answer:
The minimum distance in which the car will stop is
x=167.38m
Explanation:
[tex]39.7\frac{mi}{h}*\frac{1km}{0.621371mi}*\frac{1000m}{1km}*\frac{1h}{3600s}=17.747\frac{m}{s}[/tex]
∑F=m*a
∑F=u*m*g
The force of friction is the same value but in different direction of the force moving the car so it can stop so
[tex]F=m*a\\a=\frac{F}{m}\\a=\frac{u*m*g}{m}\\a=u*g\\a=0.096*-9.8\frac{m}{s^{2} }[/tex]
[tex]a=-0.9408 \frac{m}{s^{2}}[/tex]
[tex]v_{f}^{2}=v_{o}^{2}+2*a*(x_{f}-x_{o})\\v_{f}=0 \\x_{o}=0\\0=v_{o}^{2}+2*a*x_{f}\\x_{f}=\frac{v_{o}^{2}}{2*a} \\x_{f}=\frac{(-17.747\frac{m}{s})^{2}}{2*(-0.9408)} \\x_{f}=167.38m[/tex]
