Given :
u = 36 km/h = 36x (5/18) = 10 m/s
Time of reaction, t = 0.9 s
decelration, a = -5 m/s^2
v = 0 m/s
Let s1 be the distance covered by man when the object is seen by him
we have, u = s1/t
because s1 = u x t
= 10 x 0.9
= 9 m
when he applies brakes and decelerates at the rate of 5 m/s^2, the distance covered by him is s2
We have, v^2 = u^2 + 2as2
therefore 0 = u^2 - 2as2 ( therefore v=0 m/s)
therefore s2 = u^2/2a = 10^2/ (2 x 5)
therefore s2 =10 m
So, total distance covered s = s1 + s2 = 9 +10 =19 m
Hence 19 m distance covered by man before he stops.