Question

3) A driver in a 1000-kg car traveling at 24 m/s slams on the brakes and skids to a stop. If the coefticient of friction between the tires and the level road is 0 80 how long will the skid marks be? A) 37 m B) 30 m C) 34 m D 46 m

Answer 1

A) 37 m

Step-by-step explanation:

When the driver slams on the brakes, the force of friction acting on the car provides the deceleration that will cause the car to stop, according to Newton's second law:

`-\mu mg = ma`

where

`\mu = 0.80` is the coefficient of friction

m = 1000 kg is the mass of the car

g = 9.8 m/s^2

a is the deceleration

Substituting into the formula, we find the deceleration:

`a=- \mu g=-(0.80)(9.8 m/s^2)=-7.84 m/s^2`

Now we can find the length of the skid with the SUVAT equation:

`v^2 - u^2 = 2ad`

where

v = 0 is the final velocity of the car

u = 24 m/s is the initial velocity

d is the length of the skid

Substituting, we find

`d=(v^2 -u^2)/(2a)=(0-(24 m/s)^2)/(2(-7.84 m/s^2))=36.7 m \sim 37 m`