The coefficient of kinetic friction between tires and dry pavement is about 0.80. Assume that while travelling at 27m/s you lock your brakes and as a result the only horizontal force on the car is the frictional one.

How many seconds does it take you to bring your car to a stop?

a = u*g = 0.8 * (-9.8) = -7.84 m/s^2

V = Vo + a*t = 0
27 - 7.84*t = 0
-7.84t = -27
t = 3.44 s.

Well, it seems like you're in quite a sticky situation, or rather, a slippery one! With a coefficient of kinetic friction of 0.80, it means that the frictional force will be 0.80 times the normal force acting on the car. Since we're assuming that the only horizontal force on the car is the frictional one, this force will act in the opposite direction to your velocity, trying to bring you to a halt.

Now, let's get back to the question at hand. We know that the initial velocity is 27 m/s, and we want to find out how many seconds it takes to bring the car to a stop. We can use the equation:

v = u + at

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

In this case, the final velocity we're looking for is 0 m/s since we want to come to a stop. The initial velocity is 27 m/s and the acceleration can be calculated using the equation:

a = (frictional force) / (mass of the car)

Since the mass of the car is not given, we cannot calculate it. So unfortunately, I can't give you an exact answer to how long it will take for you to stop. But hey, don't be too disappointed! While waiting for the answer, we can entertain ourselves by coming up with car-related jokes. Why did the car go to therapy? Because it had too many breakdowns!

To calculate the time it takes to bring the car to a stop, we need to use Newton's second law of motion:

Fnet = ma

Where:
Fnet is the net force acting on the car
m is the mass of the car
a is the acceleration of the car

In this case, the net force is the force of friction, given by:

Ffriction = μ * N

Where:
μ is the coefficient of kinetic friction (0.80)
N is the normal force

The normal force can be calculated as:

N = mg

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

We need to find the acceleration, which is defined as the change in velocity per unit time:

a = Δv / Δt

Since the car initially has a velocity of 27 m/s and it comes to a stop, the change in velocity (Δv) is -27 m/s (negative because the velocity is decreasing). We can rearrange the equation to solve for Δt:

Δt = Δv / a

Let's calculate the time it takes to bring the car to a stop:

Step 1: Calculate the normal force.
N = mg
N = m * 9.8

Step 2: Calculate the force of friction.
Ffriction = μ * N

Step 3: Calculate the acceleration.
Fnet = ma
Fnet = Ffriction
a = Ffriction / m

Step 4: Calculate the time to stop.
Δt = Δv / a

Now, let's calculate the values step by step:

Step 1:
N = m * 9.8

Step 2:
Ffriction = 0.80 * N

Step 3:
Fnet = Ffriction
a = Ffriction / m

Step 4:
Δt = -27 m/s / a

Now, you'll need to provide the mass of the car in order to proceed with the calculations.

To find the time it takes for the car to come to a stop, we can use the equations of motion. In this case, the only horizontal force acting on the car is the frictional force, which can be calculated using the coefficient of kinetic friction.

The frictional force (F) can be found using the formula:
F = μ * N

Where:
μ is the coefficient of kinetic friction (0.80 in this case)
N is the normal force acting on the car, which is equal to the weight of the car (mg), where m is the mass of the car and g is the acceleration due to gravity (approximately 9.8 m/s²).

Since we know the maximum speed of the car (27 m/s) and it is assumed that the car is already traveling at this speed when the brakes are locked, the initial velocity (u) is 27 m/s and the final velocity (v) is 0 m/s.

The equation of motion that relates the initial velocity, final velocity, acceleration, and time is:
v = u + at

Since the car comes to a stop (v = 0 m/s), the final velocity is 0, and the equation becomes:
0 = 27 + at

Rearranging the equation to solve for time (t):
t = -27/a

To find the acceleration (a), we can use Newton's second law of motion, which states that the net force acting on an object is equal to its mass multiplied by its acceleration:
Fnet = ma

In this case, the net force is the frictional force (F), and the equation becomes:
F = ma
μ * N = ma

Substituting the value of N, we have:
μ * (mg) = ma

Simplifying, we get:
μg = a

Now we have the acceleration (a), we can substitute it back into the equation for time (t):
t = -27 / (μg)

Using the given coefficient of kinetic friction (μ = 0.80) and the acceleration due to gravity (g ≈ 9.8 m/s²), we can calculate the time it takes for the car to stop.

t = -27 / (0.80 * 9.8)

Solving this equation, we find that it takes approximately 3.47 seconds for the car to come to a stop.