Question

A 1200 kg car is traveling at 17 m/s when it needs to make a sudden stop. The driver jams on the brakes and the car slides to a stop. The coefficient of friction between the car and the road is 0.89. What is the car's acceleration during the stop?​

Answer 1

The car's acceleration during the stop, which is a deceleration due to friction, is calculated to be 8.73 m/s² using the coefficient of friction (μ = 0.89) and the acceleration due to gravity (g = 9.81 m/s²).

Step-by-step explanation:

The student is inquiring about the acceleration of a car during a sudden stop given a specific coefficient of friction. This is a physics problem involving Newton's laws of motion and the concept of friction. To solve this, the acceleration (a) can be found using the equation a = μ * g, where μ is the coefficient of friction (0.89) and g is the acceleration due to gravity (approximately 9.81 m/s2).

The calculation would be as follows:

a = μ * g
a = 0.89 * 9.81 m/s2
a = 8.73 m/s2

Since the car is coming to a stop, this acceleration will act in the opposite direction to the car's initial motion, meaning the car experiences a deceleration of 8.73 m/s2.