Answer:100.36 m
Explanation:
Given
speed of truck [tex]u=75 mi/hr\approx 33.528\ m/s[/tex]
Braking efficiency of 70 %
i.e. braking declaration will be 70 % of frictional force
[tex]\mu =0.8[/tex]
[tex]a_{max}=\mu g=0.8\times 10=8 m/s^2[/tex]
effective acceleration [tex]a_{eff}=0.7\times 8=5.6\ m/s^2[/tex]
using equation of motion
[tex]v^2-u^2=2 a s[/tex]
where v=final velocity
u=initial velocity
a=acceleration or declaration
s=displacement
here v will be zero
[tex]-(33.528)^2=2(-5.6)(s)[/tex]
[tex]s=\frac{33.528^2}{2\times 5.6}[/tex]
[tex]s=100.36 m[/tex]
