Question

if a train travelling at 36 km/hr is to be brought to rest in a distance of 400 m then it's retardation is

Answer 1

Given :

Initial velocity u = 36 Km/h = (36*5)÷18 = 180÷18 = 10m/s.

Distance s= 400m

Final velocity v = 0 m/s ( ∵ Object comes to rest)

Retardation a= ?

Proof:

By using 3rd Equation of Motion,

v²= u² + 2as

=> 0 = (10)²+ 2×a×400

=> 0= 100 + 800a

=> -100 = 800a

=> a =  -100 ÷800

=> a = -1 ÷ 8

=>a = -0.125 m/s²

Hence the retardation a= 0.125m/s².