Traveling at a speed of 18.3 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 0.540. What is the speed of the automobile after 1.33 s have elapsed? Ignore the effects of air resistance.

a = -ug = -0.54*9.8 = -5.29 m/s^2.

V = Vo + at = 18.3 - 5.29*1.33=11.26 m/s

To find the speed of the automobile after 1.33 seconds have elapsed, we can use the equations of motion.

Let's start by finding the deceleration of the automobile. The deceleration can be calculated using the formula:

a = μ * g

where:
a is the deceleration
μ is the coefficient of kinetic friction
g is the acceleration due to gravity (approximately 9.8 m/s²)

Given that the coefficient of kinetic friction is 0.540, we can substitute the values into the formula:

a = 0.540 * 9.8 = 5.292 m/s²

Now, using the equation of motion:

v = u + at

where:
v is the final velocity
u is the initial velocity
a is the acceleration
t is the time

Given that the initial velocity u is 18.3 m/s and the time t is 1.33 seconds, and the acceleration a is -5.292 m/s² (negative because it's deceleration), we can substitute the values:

v = 18.3 + (-5.292) * 1.33

Computing the equation:

v = 18.3 - 7.056

v = 11.244 m/s

Therefore, the speed of the automobile after 1.33 seconds have elapsed is 11.244 m/s.