Question

The tires of a car make 62 revolutions as the car reduces its speed uniformly from 92.0 km/h to 63.0 km/h. The tires have a diameter of 0.82 m. part a What was the angular acceleration of the tires? part b If the car continues to decelerate at this rate, how much more time is required for it to stop? part c If the car continues to decelerate at this rate, how far does it go? Find the total distance.

Answer 1

here,

the angle covered , theta1 = 62 * 2*pi = 389.36 rad

diameter of tire , d = 0.82 m

radius , r = d/2 = 0.41 m

for v1 = 92 km/h = 25.56 m/s

w1 = v1/r = 62.3 rad/s

for v2 = 63 km/h = 17.5 m/s

w2 = v2/r = 42.68 rad/s

a)

let the angular acceleration be alpha

using third equation of motion

w2^2 - w1^2 = 2 * alpha * theta

42.68^2 - 62.3^2 = 2 * alpha * 389.36

solving for alpha

alpha = - 2.64 rad/s^2

b)

let the time taken to stop be t

using first equation of motion

0 = w2 + alpha * t

0 = 42.68^2 - 2.64 * t

t = 16.14 s

the time taken is 16.14 s

c)

the total angle covered before stopping , phi = w1^2 / (2 * alpha)

phi = 62.3^2 /(2 * 2.64) = 735.09 rad

the distance travelled , s = r * phi

s = 301.4 m