Frictional force =
mg = 0.9mg
where m = mass of the man
Now to complete the distance of 50 m from rest he need to accelerate by a (say)
This acceleration 'a' is such that the force by man should not exceed firctional force otherwise he will slip, thus
F = ma = 0.9mg
or a = 0.9g = 9m/s2, is the maximum acceleration with which he can run
use
s = (1/2) a t2 (as the man starts from rest)
or t = (2s/a)1/2 , s= 50m, subtitute and get the result.
Yes while stopping also this is the minimum time required to stop.
Moderation Team
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