Question

A 1800 kg car moves along a horizontal road at speed v₀ = 18.4 m/s. The road is wet, so the static friction coefficient between the tires and the road is only µ static = 0.188 and the kinetic friction coefficient is even lower, µ kinetic = 0.1316. The acceleration of gravity is 9.8 m/s². What is the shortest possible stopping distance for the car under such conditions? (neglect the reaction time of the driver and round your answer to 4 decimal places)

Answer 1

Answer:

The shortest possible distance is  $|s| = 91.9 \ m$

Explanation:

From the question we are told that

   The mass of the car is   $m = 1800 \ kg$

    The speed along the horizontal road is  $v_o =u = 18.4 \ m/s$

     The static friction coefficient is  $\mu_s = 0.188$

      The kinetic friction coefficient is  $\mu_k = 0.1316$

Generally the static frictional force acting on the car is  mathematically represented as

          $F_f = m * g * \mu_s$

Generally the force propelling the car is mathematically represented as

        $F = m * a$

Here a is the maximum acceleration

at the point which the car stops ,

       $F = F_f$  

=> $m * g * \mu_s = ma$

=> $g * \mu_s =a$

=> $a = 9.8 * 0.188$

=> $a = 1.8424 \ m/s^2$

Generally from kinematic equation

    $v^2 = u^2 + 2as$

Here v  is the final velocity of the car and the value is zero given that the car comes to rest

So

        $0^2 = 18.4^2 + 2* 1.8424 s$

=>     $s = - \frac{18.4^2}{2 * 1.8424}$

=>     $|s| = 91.9 \ m$