Answer:
[tex]x=22.65m[/tex]
Explanation:
We have an uniformly accelerated motion, with a negative acceleration. Thus, we use the kinematic equations to calculate the distance will it take to bring the car to a stop:
[tex]v_f^2=v_0^2-2ax\\\frac{0^2-v_0^2}{-2a}=x\\x=\frac{v_0^2}{2a}[/tex]
The acceleration can be calculated using Newton's second law:
[tex]\sum F_x:F_f=ma\\\sum F_y:N=mg[/tex]
Recall that the maximum force of friction is defined as [tex]F_f=\mu N[/tex]. So, replacing this:
[tex]\mu N=ma\\\mu mg=ma\\a=\mu g\\a=0.901(9.8\frac{m}{s^2})\\a=8.83\frac{m}{s^2}[/tex]
Now, we calculate the distance:
[tex]x=\frac{v_0^2}{2a}\\x=\frac{(20\frac{m}{s})^2}{2(8.83\frac{m}{s^2})}\\x=22.65m[/tex]
