Final Answer:
The acceleration of the car while braking is -6.57 m/s². The final speed of the car after 1.2 seconds is 9.86 m/s. The distance travelled by the car to come to a full stop is 13.3 meters.
Step-by-step explanation:
To find the acceleration of the car, we'll use Newton's second law of motion: ΣF = ma, where ΣF is the sum of the forces acting on the car. The force of friction (f) can be calculated using the equation f = μN, where μ is the coefficient of friction and N is the normal force. The normal force is equal to the car's weight, which is mass times gravity: N = mg.
The force of friction opposes the motion and acts in the opposite direction of the car's velocity, leading to a negative acceleration. Substituting values: f = μN = μmg. Now, the net force can be determined by ΣF = ma = f, which gives us the acceleration, a = f / m = (μmg) / m = μg = 0.62 * 9.81 m/s² = -6.57 m/s².
Next, to find the final speed after 1.2 seconds, we'll use the equation v = u + at, where v is the final velocity, u is the initial velocity, a is the acceleration, and t is the time. Given u = 18 m/s, a = -6.57 m/s², and t = 1.2 seconds, substituting the values gives v = 18 m/s + (-6.57 m/s² * 1.2 s) = 9.86 m/s.
Finally, to find the distance travelled during deceleration until the car comes to a stop, we'll use the equation v² = u² + 2as, where v is the final velocity (0 m/s), u is the initial velocity (18 m/s), a is the acceleration (-6.57 m/s²), and s is the distance. Rearranging the equation to solve for distance, we get s = (v² - u²) / (2a) = (0 - 18²) / (2 * -6.57) = 324 / -13.14 = 13.3 meters. Therefore, the car travels 13.3 meters before coming to a full stop.