Answer:
70 revolutions
Explanation:
We can start by the time it takes for the driver to come from 22.8m/s to full rest:
[tex] t = \Delta v/a = (22.8 - 0)/2.95 = 7.73 s[/tex]
The tire angular velocity before stopping is:
[tex] \omega_0 = v/r = 22.8 / 0.2 = 114 rad/s [/tex]
Also its angular decceleration:
[tex] \alpha = a / r = 2.95/0.2 = 14.75 rad/s^2[/tex]
Using the following equation motion we can findout the angle it makes during the deceleration:
[tex]\omega^2 - \omega_0^2 = 2\alpha\Delta \theta[/tex]
where [tex]\omega[/tex] = 0 m/s is the final angular velocity of the car when it stops, [tex]\omega_0[/tex] = 114rad/s is the initial angular velocity of the car [tex]\alpha[/tex] = 14.75 rad/s2 is the deceleration of the can, and [tex]\Delta \theta[/tex] is the angular distance traveled, which we care looking for:
[tex]-114^2 = 2*(-14.75)*\Delta \theta[/tex]
[tex]\Delta \theta = 440rad[/tex] or 440/2π = 70 revelutions
