# Answer to Question #19420 in Mechanics | Relativity for Celesta

Question #19420

A boy on a bicycle, whose tire's radius is .5m, is traveling with a speed of 5m/s and begins braking. A bug clings tightly to the edge of the tire on the side. The tire coasts to a stop with angular acceleration of -.2rads/s^2. (a) How long did it take the bike, boy & bug to stop? (b) How many revolutions did the bug travel through before coming to a stop from th initial braking?

Expert's answer

initial angular speed omega =v/R = (5 m/s)/(0.5 m) = 10 rad/s;

if the braking time is T, then final speed = 0 = omega + acc*T = 10 rad/s -

(0.2 rad/s^2)*T;

T = (10 rad/s) / (0.2 rad/s^2) = 50 s.

The number of revolutions will be N = angle/(2*Pi) = (omega/2)*T = ( (10

rad/s)*(50 s)/2 ) / (2*Pi) = 39.81;

Floored downwards, it's 39 revolutions.

Answer: (a) 50 seconds, (b) 39 revolutions

if the braking time is T, then final speed = 0 = omega + acc*T = 10 rad/s -

(0.2 rad/s^2)*T;

T = (10 rad/s) / (0.2 rad/s^2) = 50 s.

The number of revolutions will be N = angle/(2*Pi) = (omega/2)*T = ( (10

rad/s)*(50 s)/2 ) / (2*Pi) = 39.81;

Floored downwards, it's 39 revolutions.

Answer: (a) 50 seconds, (b) 39 revolutions

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