Question

B. A car moving at an initial speed vi applies its brakes and skids for some distance until coming to a complete stop. If the coefficient of kinetic friction between tires and the road is µk what distance did the car skid?

Answer 1

Complete Question

The complete question is  shown on the first uploaded image  

Answer:

The distance which the car skid is  $l = \frac{v_i^2 }{2 * \mu_k * g }$

Explanation:

From the question we are told that  

     The  initial velocity of the car is  $v_i$

     The  coefficient of kinetic friction is  $\mu_k$

According to the law of energy conservation

    The initial Mechanical Energy =  The final  Mechanical Energy  

                           $M_i = M_f$  

The initial mechanical energy is  mathematically represented as  

              $M_i = KE _o + PE_e$

where KE is the initial kinetic energy which is  mathematically represented as

        $KE = \frac{1}{2} m v_i^2$

And  PE  is  the initial potential energy which is  zero given that the car is  on the ground

     Now  

           $M_f = W_{\mu}$

Where  $W_{\mu}$ is  the work which friction exerted on the car  which is  mathematically represented as

       $W_{\mu} = m* \mu_k * g * l$

Where  $l$ is the distance covered by the car before it slowed down

        $\frac{1}{2} m v_i^2 = m* \mu_k * g * l$

=>     $l = \frac{v_i^2 }{2 * \mu_k * g }$