Question

a car traveling at 28.4 m/s undergoes a constant deceleration of 1.92 m/s2 when the breaks are applied. How many revolutions does each tire make before the car comes to a stop

Answer 1

Complete Question

a car traveling at 28.4 m/s undergoes a constant deceleration of 1.92 m/s2 when the breaks are applied. How many revolutions does each tire make before the car comes to a stop? Assume that the car does not skid and that each tire has a radius of 0.307 m. Answer in units of rev.

Answer:

The value is  $N = 109 \ rev$      

Explanation:

From the question we are told that

    The speed of the car is  $u = 28.4 \ m/s$

     The constant deceleration experienced is  $a = 1.92 \ m/s^2$

      The radius of the tire is  $r = 0.307 \ m$

     

Generally from kinematic equation we have that

      $v^2 = u^2 + 2as$

Here  v is the final velocity which is  0 m/s

   So

         $0^2 = 28.4^2 + 2 * 1.92 * s$

=>      $s = 210.04 \ m$

Generally the circumference of the tire is mathematically represented as

         $C = 2 \pi r$

=>      $C = 2 * 3.142 * 0.307$    

=>      $C = 1.929 \ m$

Generally the number of revolution is mathematically represented as

         $N = \frac{ s}{C}$    

=>     $N = \frac{210.04}{1.929}$

=>     $N = 109 \ rev$