Answer:30.50 m
Explanation:
Given
coefficient of friction between railroad and crates [tex]\mu =0.48[/tex]
Train initial velocity (u)[tex]=61 kmph \approx 16.94 m/s[/tex]
To get the shortest distance brakes applied should be order of friction force between crates and railroad floor
[tex]a=\mu g=0.48\times 9.8=4.704 m/s^2[/tex]
using [tex]v^2-u^2=2as[/tex]
where v=final velocity
u=initial velocity
a=acceleration or deceleration
s=distance
here v=0 , u=16.94 m/s
[tex]0-(16.94)^2=2\times (-4.704)\times s[/tex]
[tex]s=\frac{(16.94)^2}{2\times 4.704}[/tex]
s=30.502 m
