Question

A train was moving at a rate of 36 km/h. When the brakes were applied. It comes to rest at a distance of 200m. Calculate the retardation produced in the train?

Answer 1

u=36km/h = 36x5/18 = 10ms-1
s= 200m
v=0
a=?

v^2-u^2 = 2as
0-100=400a
-1/4=a
a = -0.25ms^-2
Retardation is 0.25ms^-2

Answer 2

Distance covered before coming to rest =S=200 m.
Initial velocity =36km/h =36×5/18 m/s =10m/s =u.
Final velocity =v= 0m/s. Formula required =>
${v}^{2} - {u}^{2} = 2as$
=> 0-100 =2(a)(200)=> a= -100/400 = -0.25 m/s^2 . So 0.25 m/s^2 is thus the retardation produced in the train .