Answer to Question #200894 in Physics for Joel Katu

Question

During a very quick stop, a car decelerates at 7.00 m/s² .

a). What is the angular acceleration of its 0.280 m radius tires, assuming they do not slip on the pavement?

b). How many revolutions do the tires make before coming to rest, given their initial velocity is 95.0 rad/s?

c). How long does the car take to stop completely?

d). What distance does the car travel in this time?

e). What was the car's initial velocity?

f). Do the values obtained seem reasonable, considering that this stop happens very quickly?

Answer 1

a) The angular acceleration is

$$\alpha=\frac ar=25\text{ rad/s}^2.$$

b) Find the number of revolutions:

$$n=\frac{\phi}{2\pi}=\frac{\omega^2/2\alpha}{2\pi}=\frac{\omega^2}{4\alpha\pi},\\\space\\ n=28.$$

c) Determine the time:

$$t=\frac\omega\alpha=3.8\text{ s}.$$

d) The initial velocity was

$$v=\omega r=26.6\text{ m/s}.$$

e) Yes, although it was a harsh stop because the car stopped from 26.6 m/s to 0 in less than 4 s and it took 3.8 m to stop.