Question

The tires of a car make 56 revolutions as the car reduces its speed uniformly from 93 km/h to 50 km/h . The tires have a diameter of 0.76 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?

Answer 1

(A) = 56 revolutions = 56*2*pi = 351.8 radians

v1 = 93 km/h = (93 * 1000) / (60*60) = 25.83 m/s

So, w1 = v1 / r = 25.83/0.38 = 68 rad/s

v2 = 50 km/h = (50*1000) / (60*60) = 13.89 m/s

=> w2 = 13.89/0.38 = 36.5 rad/s

Use the expression -

w2^2 = w1^1 + 2**

=> 36.5^2 = 68^2 + 2**351.8

=> = (36.5^2 - 68^2) / (2*351.8) = -4.68 rad/s^2

So, the magnitude of the angular acceleration = 4.68 rad/s^2

(B) When the car stop, w2 = 0

use the expression -

w2 = w1 + *t

=> 0 = 68 + (-4.68)*t

=> t = 68 / 4.68 = 14.5 sec.

So time of stopping of the car = 14.5 s.