Answer:
86.14 meters.
Explanation:
Step one:
Given data
velocity of car = 26 m/s
the coefficient of static friction between the tires and the road
µ = 0.4 (kinetic)
Let us take g = 9.81 m/s^2
Required
The distance x = distance in m
We know that
[tex]N = Fg = mg\\\\F_\mu = -\mu N = \mu F_g = \mu mg\\\\KE = (1/2) mv^2[/tex]
W = F*x (Work is force times distance)
Step two:
Conservation of energy gives
KE = W
Substituting gives
[tex](1/2) mv^2 = F \mu x\\\\(1/2) mv^2 = \mu mgx\\\\mv^2 = 2 \mu mgx[/tex]
Solving for distance (x) gives
[tex]x = mv^2 / 2 \mu mg[/tex]
Simplifying
[tex]x = v^2 / 2 \mu g[/tex]
Substitute:
[tex]x = v^2 / 2 \mu g[/tex]
[tex]x= 26^2/2*0.4*9.81[/tex]
[tex]x=676/7.848\\\\x=86.14[/tex]
Therefore, the minimum braking distance is 86.14 meters.
