A car is traveling at 51.8 km/h on a flat highway. (a) If the coefficient of friction between the road and the tires on a rainy day is 0.18

The deceleration force, F,  is F = 0.183*mass*g (g is acceleration of gravity).  This equals -mass*a, where 'a' is acceleration (deceleration) of the car, so

a = -0.183*9.81m/s². (The negative sign just means it is slowing down.)

With constant acceleration we have the kinetic equation v² = u² +2as, where v = final velocity (0), u = initial velocity (51.8km/hr = 51.8*10³/3600 m/s) and s is distance travelled.  So:

0 = (51.8*10³/3600)² - 2*0.183*9.81*s

 s = (51.8*10³/3600)²/2*0.183*9.81

$${\mathtt{s}} = {\frac{{\left({\frac{{\mathtt{51.8}}{\mathtt{\,\times\,}}{{\mathtt{10}}}^{{\mathtt{3}}}}{{\mathtt{3\,600}}}}\right)}^{{\mathtt{2}}}}{\left({\mathtt{2}}{\mathtt{\,\times\,}}{\mathtt{0.183}}{\mathtt{\,\times\,}}{\mathtt{9.81}}\right)}} \Rightarrow {\mathtt{s}} = {\mathtt{57.663\: \!954\: \!885\: \!109\: \!463\: \!3}}$$  metres

or s ≈ 57.7 metres