Answer:0.43
Explanation:
Given
mass of car [tex]m=2400 kg[/tex]
Speed of car [tex]u=14 m/s[/tex]
Distance traveled before coming to halt [tex]s=23.2 m[/tex]
Let [tex]\mu [/tex]the coefficient of friction
Maximum deceleration road can provide during motion is
[tex]a=\mu g[/tex]
using [tex]v^2-u^2=2 as [/tex]
[tex]0-14^2=2\cdot (-\mu g)\cdot 23.2[/tex]
[tex]\mu =\frac{196}{454.72}[/tex]
[tex]\mu =0.431[/tex]