A car of mass 1000.0 kg is traveling along a level road at 100.0 km/h when its brakes are applied. Calculate the stopping distance if the coefficient of kinetic friction of the tires is 0.500. Neglect air resistance. (Hint: since the distance traveled is of interest rather than the time, xx is the desired independent variable and not tt. Use the Chain Rule to change the variable: \frac{dv}{dt}=\frac{dv}{dx}\frac{dx }{dt}=v\frac{dv}{dx} dt dv = dx dv dt dx =v dx dv .)

Explanation:

Given:

Now the kinetic frictional force:

where:

N= normal force of reaction by the ground on the car, which is equal to the weight of the car

Therefore the acceleration of the car due to frictional force:

Now the stopping distance:

this is an example of "committed" time.

committed time, as contracted time, takes need over essential and available time since it is seen as profitable work. it alludes to the time allotted to keep up a home and family. at the point when a man is commuting using committed time this individual may feel that the drive is an obligation to family, for example, strolling youngsters to class or driving a mate to work. contracted and committed time clients may feel that their commute could really compare to the commute of important or leisure time clients on the grounds that their commute is profitable work. in this manner, they might be more disposed to pick a mechanized method of movement.  

conservation of mass

mass before something happens = mass afterwards