Question

The driver of a car traveling at 22.8 m/s applies the brakes and undergoes a constant deceleration of 2.95 m/s 2 . How many revolutions does each tire make before the car comes to a stop, assuming that the car does not skid and that the tires have radii of 0.2 m? Answer in units of rev.

Answer 1

Answer:

70 revolutions

Explanation:

We can start by the time it takes for the driver to come from 22.8m/s to full rest:

$t = \Delta v/a = (22.8 - 0)/2.95 = 7.73 s$

The tire angular velocity before stopping is:

$\omega_0 = v/r = 22.8 / 0.2 = 114 rad/s$

Also its angular decceleration:

$\alpha = a / r = 2.95/0.2 = 14.75 rad/s^2$

Using the following equation motion we can findout the angle it makes during the deceleration:

$\omega^2 - \omega_0^2 = 2\alpha\Delta \theta$

where $\omega$ = 0 m/s is the final angular velocity of the car when it stops, $\omega_0$ = 114rad/s is the initial angular velocity of the car $\alpha$ = 14.75 rad/s2 is the deceleration of the can, and $\Delta \theta$ is the angular distance traveled, which we care looking for:

$-114^2 = 2*(-14.75)*\Delta \theta$

$\Delta \theta = 440rad$ or 440/2π = 70 revelutions