a) The wheel of the car decelerates at an angular acceleration of - 0.577 radians per square second.
b) The wheel of the car requires a time of 51.149 seconds to stop.
c) The wheel of the car travels a distance of 683.225 meters before stopping.
How to analyze a decelerating rotating wheel
a) In this case we have a rotating wheel that decelerates at constant rate. The angular acceleration, in radians per square second, of the tires is determined by the following formula:
α = [ω'² - ω²] / (2 · θ) (1)
Where:
- ω - Initial angular velocity, in radians per second.
- ω' - Final angular velocity, in radians per second.
- θ - Change in the angular displacement, in radians.
The initial and final angular velocities, in radians per second, are now determined:
ω' = v' / R (2)
ω = v / R (3)
Where:
- v - Initial linear velocity, in meters per second.
- v' - Final linear velocity, in meters per second.
- R - Radius of the tire, in meters.
If we know that R = 0.80 m, v = 23.611 m / s, v' = 15.556 m / s and θ ≈ 427.257 radians, then the angular acceleration of the tire is:
ω' = (15.556 m / s) / (0.80 m)
ω' = 19.445 rad / s
ω = (23.611 m / s) / (0.80 m)
ω = 29.513 rad / s
α = [(19.445 rad / s)² - (29.513 rad / s)²] / [2 · (427.257 rad)]
α = - 0.577 rad / s²
b) The time required to stop the car, in seconds, is determined by the following expression:
t = (ω' - ω) / α (4)
t = (0 rad / s - 29.513 rad / s) / (- 0.577 rad / s²)
t = 51.149 s.
c) First, we find the change in angular displacement of the tire:
θ = [ω'² - ω²] / (2 · α) (5)
θ = [(0 rad / s)² - (29.513 rad / s)²] / [2 · (- 0.577 rad / s²)]
θ = 754.781 rad
Lastly, the distance traveled by the vehicle is:
s = R · θ (6)
s = (0.80 m) · (754.781 rad)
s = 683.825 m
To learn more on angular motion: https://brainly.com/question/14979994
#SPJ1
