43.8 m
Explanation:
The only force acting on the car is the frictional force f and since friction is a force that is always directed opposite to the direction of motion, it will have a negative sign. Therefore, we can write Newton's 2nd law as
[tex]x:\;\;\;\;F_{net} = -f = -\mu N = ma[/tex] (1)
[tex]y:\;\;\;\;F_{net} = N - mg = 0[/tex] (2)
Substituting Eqn(2) into Eqn(1), we get
[tex]\Rightarrow a = -\mu g[/tex] (3)
To find the stopping distance x, we are going to use the equation
[tex]v^2 = v_0^2 + 2ax[/tex]
Initially, the car was moving at [tex]v_0 = 27\:\text{m/s}.[/tex] When it stopped, its final velocity [tex]v[/tex] is zero. Using these, as well as Eqn(3) on the equation for v^2, we find that
[tex]0 = v_0^2 + 2(-\mu g)x \Rightarrow 0 = v_0^2 - 2\mu gx[/tex]
or solving for x,
[tex]x = \dfrac{v_0^2}{2\mu g} = \dfrac{(27\:\text{m/s})^2}{2(0.85)(9.8\:\text{m/s}^2)}[/tex]
[tex]\:\:\:\:= 43.8\:\text{m}[/tex]
