solution:
maximum acceleration produce by bag =60 g
time to come to stop = 36 ms
applying equation
[tex]V=u-at[/tex]
inserting values
[tex]0=u-(60)(36*10^-^3)[/tex]
[tex]u=(600*36)/1000=21.6 m/s\\[/tex]
now to find distance of penctration:
[tex]S=ut-1/2at^2[/tex]
inserting values
[tex]S= (21.6)(36*10^-^3)-1/2(600)(0.036)^2[/tex]
[tex]S=0.38m[/tex]
hence distance traveled by person before coming to rest is 0.38 m
