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 angular 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?

Solution 8PE

From the data given, we are required to calculate various quantities considering the equations of rotational motion.

Part a

Step 1 of 6</p>

The rate of deceleration of the car is m/s2

The radius of the tire is m

Therefore, the angular acceleration

rad/s2

rad/s2

Therefore, the angular acceleration of the tire is 25.0 rad/s2.

Part b

Step 2 of 6</p>

Since the tires come to rest, hence final angular velocity is .

Initial angular velocity rad/s

Therefore, the time taken

s

s

The angular displacement of the tires at 3.80 s is,

rad

rad

Now, rev = rad

So, 180.5 rad rev

rad rev

rev

The tires would make approximately 29 revolutions before coming to rest.

Part c

Step 3 of 6</p>

Since the tires come to rest, hence final angular velocity is .

Initial angular velocity rad/s

Therefore, the time taken

s

s

Therefore, the car would take 3.80 s to stop completely.

Part d