Question

The tires of a car make 74 revolutions as the car reduces its speed uniformly from 86 km/h to 43 km/h . The tires have a diameter of 0.81 m .What was the angular acceleration of the tires?If the car continues to decelerate at this rate, how much more time is required for it to stop?

Answer 1

Solution:-

Given –

Vf = 43/3.7 = 11.62m/s

Vi = 86/3.7 = 23.24m/s

d = 74/0.81*3.14

d = 188.21m

Δt = 2d/(Vf+Vi)

Δt = 2*188.21/(11.62+23.24)

Δt = 376.4/34.86

Δt = 10.79 sec.

a = Δv/Δt = (23.24-11.62)/10.79

a = 11.62/10.79

a = 1.076 m/s²

a)      Calculate angular acceleration

ω = a/r = 1.076/0.405 = 2.656 rad/s²

ω = 2.656 rad/s²

b)     If the car continuous to decelerate at this rate how much more times is required for it to step

ts = Vf /a

ts = 11.62m/s / 1.076m/s²

ts = 10.79 sec