A car without ABS (antilock brake system) was moving at \(15.0 \mathrm{~m} / \mathrm{s}\) when the driver hit the brake to make a sudden stop. The coefficients of static and kinetic friction between the tires and the road are 0.550 and 0.430 , respectively. a) What was the acceleration of the car during the interval between braking and stopping? b) How far did the car travel before it stopped?

First, we need to find the kinetic friction force acting on the car when the driver hits the brake. The formula for kinetic friction force (\(F_k\)) is given by: \(F_k = \mu_k F_N\) where \(\mu_k\) is the coefficient of kinetic friction and \(F_N\) is the normal force, which is equal to the car's weight in this case. Since the car is on a horizontal road, the normal force (\(F_N\)) is equal to the gravitational force, which can be calculated as: \(F_N = mg\) where \(m\) is the mass of the car and \(g\) is the acceleration due to gravity, approximately 9.81 \(\mathrm{m/s^2}\). We are not given the mass of the car, but we can express the friction force in terms of mass and then find the acceleration.

Using Newton's second law of motion, we have: \(F = ma\) where \(F\) is the net force acting on the car, \(m\) is the mass of the car, and \(a\) is the acceleration. Since the net force acting on the car is the kinetic friction force, we have: \(ma = F_k\) Substituting the expression for kinetic friction force from Step 1, we get: \(ma = \mu_k mg\) We can then solve for the acceleration (\(a\)): \(a = \mu_k g\) Insert the given values for the coefficient of kinetic friction and the acceleration due to gravity: \(a = 0.430 \times 9.81\) \(a = -4.22 \mathrm{~m/s^2}\) (The negative sign indicates the acceleration is opposite to the initial velocity, as it's a decelerating motion) The acceleration of the car during the interval between braking and stopping is \(-4.22 \mathrm{~m/s^2}\).

To find the distance the car traveled before stopping, we can use the following kinematic equation: \(V_f^2 = V_i^2 + 2as\) Here, \(V_f\) is the final velocity, which is 0 since the car stopped, \(V_i\) is the initial velocity, which is \(15.0 \mathrm{~m/s}\), \(a\) is the acceleration we found in Step 2, which is \(-4.22 \mathrm{~m/s^2}\), and \(s\) is the stopping distance we need to find. Plug in the given values and solve for \(s\): \(0 = (15.0)^2 + 2(-4.22)s\) Solve for \(s\): \(s = \frac{(15.0)^2}{2 \times 4.22}\) \(s \approx 26.89 \mathrm{~m}\) The car traveled approximately \(26.89 \mathrm{~m}\) before it stopped.