A car is traveling at 50.8 km/h on a flat highway.?

[DeannaDior]

If the coefficient of friction between the road and the tires on a rainy day is 0.100, what is the minimum distance needed for the car to stop?

What is the stopping distance when the surface is dry and the coefficient of friction is 0.600?

 

[brian]

In m/s:
50.8 km/h
= (50.8 km x 1000 m/km)/(1 h x 3600 s/h)
= 50800/3600 m/s
= 14.11 m/s.

Then, we have:

F(net) = F(car) - F(friction)
==> ma = mg - u*mg
==> a = g - u*g
==> a = (1 - u)g
= (1 - 0.100)(-9.81 m/s²)
= -8.829 m/s².
(This is the acceleration needed to stop)

Then, the distance required to stop is:

v?² = v?² + 2ad
==> d = (v?² - v?²)/(2a)
= [(0.000 m/s)² - (14.11 m/s)²]/[2(-8.829 m/s²)]
= 11.3 m.

You can re-work this with u = 0.600 to get the answer to the second part.

I hope this helps!