Physics: a 2500kg car is travelling at a constant speed of 14 m/s along an icy, but straight and level road. the driver of the car, seeing an approaching traffic light turn red., slams on the brakes. Wheels locked and ttired skidding, the car slide to a halt in a distance of 25m. What is the coefficient of friction

I think its half of pie so you divid that by 3.14 ....

a 2500kg car is travelling at a constant speed of - id250017077826, carrietaylor234

byers024

30.01.2022

0.40

Explanation:

Given that :

the mass of the caer = 250 kg

initial speed = 14 m/s

final speed = 0 m/s

distance s = 25 m

Using the equation of motion

v^2 = u^2 + 2as

making a the subject of the formula ; we have:

v^2-u^2 = 2as

$a= \dfrac{v^2-u^2 }{2 \ s}$

$a= \dfrac{(0)^2-(14)^2 }{2 \ (25)}$

$a= \dfrac{0-196 }{50}$

$a= \dfrac{-196 }{50}$

a = -3.92 m/s²

However; the relation for the coefficient of the kinetic static friction can be expressed as:

$f= \mu_k *mg= ma$

$f= \mu_k *g= a$

$f= \mu_k = \dfrac{a}{g}$

$f= \mu_k = \dfrac{3.92}{9.8}$

$f= \mu_k = 0.40$

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