The tires of a car make 90 revolutions as the car reduces its speed uniformly from 88.0 km/hkm/h to 56.0 km/hkm/h. The tires have a diameter of 0.84 mm. 1. What was the angular acceleration of the tires?
2. If the car continues to decelerate at this rate, how much more time is required for it to stop?
3. If the car continues to decelerate at this rate, how far does it go? Find the total distance.

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

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

the moon affects the earth gravitation ally, causing certain motions of that body and also producing tides in its ocean and atmosphere. the tidal wave in our atmosphere is, however, far too minute to be an effective cause in any known meteorological phenomenon.

explanation: