If the coefficient of kinetic friction between tires and dry pavement is 0.800, what is the shortest distance in which an automobile can be stopped by locking the brakes when traveling at 29.4 m/s ?

The shortest distance in which an automobile can be stopped by locking the brakes when traveling at 29.4 m/s = 55.125 m = 55.13 m

Explanation:

Coefficient of kinetic friction = 0.8

Frictional force can be evaluated using μN

where N is the normal reaction = mg

Fr = 0.8 m × 9.8 = 7.84 m

The car stops with th help of friction, this friction manifests in a form of decelerating force.

This frictional force will be equal to the ma force,

ma = 7.84 m

a = 7.84 m/s²

But Since it's deceleration, this acceleration is - 7.84 m/s²

So, using the equations of motion,

u = initial velocity = 29.4 m/s

v = final velocity = 0 m/s (since it comes to a stop finally)

x = distance travelled before stopping

a = - 7.84 m/s²

v² = u² + 2ax

0² = 29.4² + 2(-7.84)(x)

15.68 x = 864.36

x = 55.125 m = 55.13 m

where’s the list ? the question is incomplete