Consult Multiple Concept Example 10 in preparation for this problem. Traveling at a speed of 15.4 m/s, the driver of an automobile suddenly locks the wheels by slamming on the brakes. The coefficient of kinetic friction between the tires and the road is 0.680. What is the speed of the automobile after 1.11 s have elapsed? Ignore the effects of air resistance.

Answer: 8.07 m/s

Explanation:

Applying the definition of acceleration, as the rate of change of velocity, we can extract the value of the velocity v at a given time t, as follows:

v = v₀ + a t (1)

As we already know v₀ and t, our single unknown is the acceleration a.

Once the friver slammed on the brakes, there exists an external force, due to friction, which is the only external force acting in the horizontal direction, producing a deceleration.

This friction force, is equal to the product of the coeficient of kinetic friction, times the normal force.

In this case, the normal force is numerically equal to the gravity force, which we call weight.

This force opposes to the direction of the initial velocity, as it always opposes to the relative movement between both surfaces in contact.

Applying Newton's 2nd Law, we get:

Fnet = ma ⇒ Ff = m a ⇒ - μk. m. g = m. a

Solving for a:

a = - μk. g = - 0.68. 9.8 m/s² = - 6.66 m/s²

Replacing in (1)

v = 15.4 m/s + (-6.66) m/s². 1.11 s = 8.07 m/s