Question

Rafael is driving his car at 26 m/s. What is the shortest distance in which he can brake and stop if the coefficient of static friction between the tires and the road is 0.4

Answer 1

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

$N = Fg = mg\\\\F_\mu = -\mu N = \mu F_g = \mu mg\\\\KE = (1/2) mv^2$

W = F*x  (Work is force times distance)

Step two:

Conservation of energy gives  

KE = W

Substituting gives  

$(1/2) mv^2 = F \mu x\\\\(1/2) mv^2 = \mu mgx\\\\mv^2 = 2 \mu mgx$

Solving for distance (x) gives  

$x = mv^2 / 2 \mu mg$

Simplifying

$x = v^2 / 2 \mu g$

Substitute:  

$x = v^2 / 2 \mu g$

$x= 26^2/2*0.4*9.81$

$x=676/7.848\\\\x=86.14$

Therefore, the minimum braking distance is 86.14 meters.