Traveling at a speed of 15.5 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 0.640. What is the speed of the automobile after 1.45 s have elapsed? Ignore the effects of air resistance.

vf=vi-at but a=force/mass=mu(mg)/m=mu*g

solve for Vf

To find the speed of the automobile after 1.45 seconds have elapsed, we can use the equation:

v = u + at

Where:
v = final velocity
u = initial velocity
a = acceleration
t = time

In this case, the initial velocity u is given as 15.5 m/s. To find the acceleration, we need to first calculate the force of kinetic friction acting on the automobile.

The force of kinetic friction can be calculated using the equation:

Fk = μk * N

Where:
Fk = force of kinetic friction
μk = coefficient of kinetic friction
N = normal force

The normal force N is equal to the weight of the automobile, which can be calculated using the equation:

N = mg

Where:
m = mass of the automobile
g = acceleration due to gravity (approximately 9.8 m/s^2)

Since the mass of the automobile is not given, it is not possible to calculate the exact normal force. However, we can assume that the automobile is on a level surface, so the normal force is equal to the weight of the car.

Now, using the force of kinetic friction, we can calculate the acceleration using Newton's second law:

F = ma

Substituting the force of kinetic friction for F, we get:

Fk = ma

μk * N = ma

μk * mg = ma

μk * g = a

Substituting the values for μk and g, we find:

a = 0.640 * 9.8

a ≈ 6.272 m/s^2

Now that we have the acceleration, we can use the equation v = u + at to find the final velocity:

v = 15.5 + (6.272 * 1.45)

v ≈ 24.28 m/s

Therefore, the speed of the automobile after 1.45 seconds have elapsed is approximately 24.28 m/s.