Question

A car is traveling at 58.2 km/h on a flat highway. The acceleration of gravity is 9.81 m/s 2 . a) If the coefficient of kinetic friction between the road and the tires on a rainy day is 0.178, what is the minimum distance needed for the car to stop? Answer in units of m.\

Answer 1

The answer is 75.4 meters

Step-by-step explanation:

The car is traveling with constant velocity, when the car stop the fritcion force will reduce the velocity to zero, so:

`Fr=-uk*mg\\Fr=0.178*mc*9.81=mc*1.74N\\ar=(mc*1.74N)/(mc)\\ar=1.74m/s^2`

so using the accelerated motion formulas:

`vf^2=vo^2+a*d\\`

`vo=58.2km/h*(1h)/(3600s)*(1000m)/(1km)=16.2m/s`

`0=(16.2m/s)^2-2*(1.74m/s^2)*d\\d=75.4m`