Question

The driver of a car traveling at 17.7 m/s ap-

plies the brakes and undergoes a constant
deceleration of 1.38 m/s.
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 0.42 m?
Answer in units of rev.

Answer 1

43 revolutions.

Step-by-step explanation:

The time it takes for the car to stop is

`t=v_i/a=(17.7m/s)/(1.38m/s^2)=12.83s`

The distance it travels in that time is

`d=(1)/(2)at^2=(1)/(2)(1.38)(12.83) ^2=113.5m`

The number of revolutions that the tires make is then

`n=(d)/(2\pi r) =(113.5)/(2\pi *0.42) =\boxed{43\:revolutions}`

The last calculation just asks the question "how many tire circumferences fit into d=113 meters?"