Question

A train travels 30m/s then breaks are suddenly applied producing a retarding force equal to 1/10 of the weight of the train runs on the distance through which the train runs on the track before coming to rest in.

Answer 1

F = - mg/10 = m a
so
a = -g/10

v = Vi + at = 30 - g t /10
at stop v = 0
t = 300/g = time to stop

d = Vi t + (1/2)a t^2
= 30(300/g) -(1/2)(g/10)(300^2/g^2)
= 9000/g - (1/2)(9000/g)
= 4500/g
about 450 meters

Answer 2

alternate way using energy
(1/2) m Vi^2 = F d
F = mg/10
(1/2) m Vi^2 = m g d/10
(30)^2 = g d/5
900 *5 /g = d
d = 4500/g