Answer :
Answer:0.3
Explanation:
Given
velocity of car=15 m/s
truck brought to halt in a distance of 38 m
We know
[tex]v^2-u^2=2as[/tex]
Final velocity (v)=0
[tex]0-(15)^2=2(a)(38)[/tex]
[tex]a=\frac{-225}{76}[/tex]
[tex]a=-2.96 m/s^2[/tex] (deceleration)
Therefore minimum coefficient of friction \mu will be
[tex]\mu \times g=a[/tex]
[tex]\mu =\frac{a}{g}[/tex]
[tex]\mu =\frac{2.96}{9.8}=0.302[/tex]