g The tires of a car make 75 revolutions as the car reduces its speed uniformly from 91 km/h to 48 km/h . The tires have a diameter of 0.78 m . Part A What was the angular acceleration of the tires

   α = -0.01625 rad / s²

Explanation:

This is an exercise in angular kinematics, we can use the relation

         w = w₀ + 2 α θ

linear and angular variables are related

        v = w r

         w = v / r

Let's reduce the magnitudes to the SI system

         v₀ = 91 km / h (1000m / 1km) (1h / 3600s) = 25.278 m / s

         v = 48 km / h = 13,333 m / s

         θ  = 75 rev (2π rad / 1 rev) = 471.24 rad

Let's find the angular velocities

         w₀ = v₀ / r

         w₀ = 25.278 / 0.78

         w₀ = 32,408 rad / s

         w = v / r

         w = 13.333 / 0.78

         w = 17.09 rad / s

we calculate the angular acceleration

         α = (w- w₀) / 2θ  

         α = (17.09 - 32.408) / (2 471.24)

         α = -0.01625 rad / s²

the negative sign indicates that the wheel is stopping

wavelength divides in half and speed stays the same.