A driver sees a horse on the road and applies the brakes so hard that they lock and the car skids to a stop in 24 m. the road is level and the coefficient of kinetic friction between the tires and the road is 0.7. How fast was the car going when the brakes were applied

18.15m/s

Explanation:

Recall for coefficient of kinetic friction

F=UR

We're F=m*a= U*m*g

Hence a= U*g

We're a= acceleration,U coefficient of kinetic friction,g= acceleration due to gravity hence a=24*0.7=6.867m/s^2 because break was applied it is a negative acceleration hence a= -6.867m/s^2

But V=v+at at final velocity V=0

Hence 6.867*T=v T=v/6.867

But average speed =v/2

Using Newton law of motion

S=(V+v)T/2

Substitute for S,V,v and T

24=(v/2)v/6.867

v^2=329.6

V=18.15m/s

answer: option c.

explanation:

where is the circuit diagram ?