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

86.14 meters.

Step-by-step 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.