Answer:
1) α = 2.2 rad / s² , 2) t = 7.068 s , 3) in this interval s = 23.096 m
total distance s = 57.4 m
Step-by-step explanation:
For this exercise we use the angular kinematic relations, before starting the problem we reduce all the magnitude to the SI system
v₀ = 88 km / h (1000 m / 1km) 1h / 3600s) = 24.44 m / s
v = 56.0 km / h = 15.55 m / s
θ = 90 rev (2pi rad / 1 rev) = 565,487 rad
d = 0.84 m
r = d / 2
r = 0.84 / 2
r = 0.42 m
1) ask for angular acceleration
w² = w₀² - 2 α Δθ
α = (w₀² -w²) / 2 Δθ
To find the angular velocities we use the acceleration between the linear and angular velocity
v = w r
w = v / r
w₀ = 24.44 / 0.42
w₀ = 58.20 rad / s
w = 15.55 / 0.42
w = 30.037 rad / s
we calculate
α = (58.20² - 30.037²) / (2 565.487)
α = 2.2 rad / s²
2) how much longer does it take to stop
w₂ = 15.55 rad / s
w = 0
w = w₂ - α t
t = (w₂ -0) / α
t = 15.55 / 2.2
t = 7.068 s
3) the distance that the car travels from the beginning of the movement, we can find it by looking for the number of revolutions until it stops and then using the relationship between the angular and linear variable
w² = w₀² - 2 α θ
at the end of the movement speed is zero
0 = w₀² - 2 α θ
θ = w₀² / 2 α
θ = 24.44² / (2 2.2)
θ = 135.80 rad
If the angles are measured in radians, we can apply the relation
θ = s / R
s = R ttea
s = 0.42 135
s = 57.4 m
this is the distance from when the movement starts
the distance for the final part of the movement is
w = 15.55 rad / s
θ = w² / 2 α
θ = 15.55 2 / (2 2.2)
θ = 54.99 rad
the distance in this interval is
s = 0.42 54.99
s = 23.096 m