Question

The tires of a car make 62 revolutions as the car reduces its speed uniformly from 95.0 km/h to 59.0 km/h . The tires have a diameter of 0.84 m. A) What was the angular acceleration of the tires? B) If the car continues to decelerate at this rate, how much more time is required for it to stop? C) How far does the car go? Find the total distance.

Answer #1

Here ,

theta = 62 revs

radius ,r = 0.84/4 = 0.42 m

initial speed ,u = 95 km/hr = 26.4 m/s

final speed , v = 59 km/hr = 16.4 m/s

a) let the angular acceleration is a

Using third equation of motion

(16.4/0.42)^2 - (26.4/0.42)^2 = 2 * a * 62 * 2pi

a = -3.11 rad/s^2

the angular acceleration is -3.11 rad/s^2

B) let the time taken is t

t * 3.11 = (16.4/0.42)

t = 12.6 s

the time taken is 12.6 s

C)

for the total distance

(26.4)^2 = 2 * 3.11 * 0.42 * d

**d = 263.9 m**

**the total distance is 263.9 m**

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