Question

A train running at 108 km per brought to a half in 2 minutes calculated the retardation produced by the application of brakes by also calculate the distance the train travel before stopping

Answer -0.25m/s square , 1800m

Answer 1

Initial velocity(u) = 108 km/h

= 108 × 5/18
= 30m/s

Time taken (t) = 2min = 120second

Final velocity(v) = 0

Retardation = (0 - u)/t

= (0 - 30)/120

= -30/120

= -1/4

= -0.25m/s

Distance travelled before stopping =>

2as = v² - u²

2×(-0.25)×(s) = (0)² - (30)²

-0.50s = -900

s = -900/-0.5

s = 1800m

Therefore distance travelled by train before stopping = 1800m

Answer 2

I think the word is halt and not half

I take it as halt so final velocity is 0  (v)

Initial  velocity (u) = 108km/ h  = 30m/ s ( × 5/18)

a = (v - u )/t = 0-30/ 120 = -0.20 m s-2 

Since acceleration is -0.25 ms-2 therefore retardation is 0.25ms-2

v2 = u2 + 2as
900= 2as

2S =  -900 / -0.25 = > 
  S = 3600/2  = 1800 m 

Hope it helps

Thank u★★★
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