Traveling at a speed of 12.1 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 0.740. What is the speed of the automobile after 1.42 s have elapsed? Ignore the effects of air resistance.
To find the speed of the automobile after 1.42 seconds have elapsed, we can use the equation:
vf = vi + at
Where:
- vf represents the final velocity (speed) of the automobile
- vi represents the initial velocity (speed) of the automobile
- a represents the acceleration
- t represents the time elapsed
In this case, the initial velocity (vi) is 12.1 m/s (given in the question), the acceleration (a) can be calculated using the coefficient of kinetic friction (μ), and the time elapsed (t) is 1.42 seconds (also given in the question).
To calculate the acceleration (a), we can use the formula:
a = μ * g
Where:
- μ represents the coefficient of kinetic friction
- g represents the acceleration due to gravity (approximately 9.8 m/s²)
In this case, the coefficient of kinetic friction (μ) is given as 0.740.
First, let's calculate the acceleration:
a = 0.740 * 9.8 m/s²
a ≈ 7.252 m/s²
Now, we can plug the values into the first equation to find the final velocity (vf):
vf = 12.1 m/s + (7.252 m/s² * 1.42 s)
vf ≈ 22.518 m/s
Therefore, the speed of the automobile after 1.42 seconds have elapsed is approximately 22.518 m/s.