Explanation:
Given data:
mass of the car = 1500 kg
maximum friction force = 7000 N
initial velocity v_i = 20 m/s ( it is not given in the question just an assumption)
final velocity v_f = 0 m/s
[tex]\begin{array}{l}
\sum F_{y}=M g-F_{n}=0 \\
\sum F_{x}=-F_{s}=m a_{x} \\
-F_{s}=m a_{x}
\end{array}[/tex]
[tex]a_{x}=\frac{-F_{s}}{m}=\frac{-7000}{1500}[/tex]
[tex]a_{x}=-4.7 \mathrm{~m} / \mathrm{s}^{2}[/tex]
Now we can find the distance from this formula:
[tex]v_{f x}^{2}=v_{i x}^{2}+2 a_{x}(\Delta x)[/tex]
[tex]0=20^{2}+(2 \times-4.7 \times \Delta x)[/tex]
[tex]20^{2}=9.4 \Delta x[/tex]
[tex]\Delta x=\frac{20^{2}}{9.4}=42.55 \mathrm{~m}[/tex]
So, the shortest distance in which the car can stop safely without kidding
=42.55 m
