If the coefficient of kinetic friction between tires and dry pavement is 0.98, what is the shortest distance in which you can stop an automobile by locking the brakes when traveling at 34.7 m/s

The shortest distance is 62.7 m

Explanation:

Given;

coefficient of kinetic friction, μk = 0.98

initial velocity, u = 34.7 m/s

Frictional force on the tire;

Fk = -μkN

where;

N is normal reaction = mg

ma = -μkN

ma = -μkmg

a = -μkg

a = - 0.98 x 9.8 = -9.604 m/s²

The shortest distance in which you can stop an automobile by locking the brakes:

Apply equation of motion;

v² = u² + 2ax

where;

v is the final velocity

u is the initial velocity

a is the acceleration of the  automobile

0 = 34.7² + 2(-9.604)x

0 = 1204.09 - 19.208x

19.208x = 1204.09

x = 1204.09/19.208

x = 62.7 m

Therefore, the shortest distance in which you can stop an automobile by locking the brakes when traveling at 34.7 m/s is 62.7 m

measure their densities same density, same sub

jupiter is the largest planet in our solar system, having a mass and radius that are respectively, 318 and 11.2 times that of earth. suppose that an object falls from rest near the surface of each planet and that the acceleration due to gravity remains constant during the fall. each object falls the same distance before striking the ground. determine the ratio of the time of fall on jupiter to that on earth.