The tires of a car make 77 revolutions as the car reduces its speed uniformly from 92.0 km/h to 60.0 km/h. The tires have a diameter of 0.84 m. 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.

The angular acceleration of the tires is -2.2 rad/s².

If the car continues to decelerate at this rate, the time required to stop is 27.66 s.

The total distance traveled by the car before stopping is 210.96 revolutions.

The given parameters;

number of revolutions of the tire, N = 77 revinitial linear speed of the car, u  = 92 km/h = 25.56 m/sfinal linear speed of the tire, v = 60 km/h = 16.67 m/sdiameter of the tire, d = 0.84 mradius of the tire, r = 0.42 m

The angular acceleration of the tire is calculated as follows;

When the car stops, the final angular speed = 0. The time for the motion is calculated as;

The total distance traveled by the car before stopping;

total distance = 133.96 + 77 = 210.96 revolutions.

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Explanation:

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