Question

A train stopping at two stations d distance apart takes time t on the journey from one station to other. Assuming that its motion is first of uniform acceleration α and then immediately of uniform retardation β, show that 1/α+1/β=t^2/2d

Answer 1

Answer:

1/α + 1/β = T²/2d

Explanation:

A train stopping at two stations d distance apart takes time t on the journey from one station to other. Assuming that its motion is first of uniform acceleration α and then immediately of uniform retardation β, show that 1/α+1/β=t^2/2d

Train starts from rest

acceleration = α

Time T₁ for acceleration

Velocity = αT₁

Time T₂ for  retardation

0 = αT₁  - βT₂

=>  αT₁ = βT₂ = V

Distance in acceleration = (0 + αT₁ )/2  * T₁ = αT₁ ²/2 = V²/2α

Distance in retardation = βT₂ ²/2 = V²/2β

Total Distance =V²/2α  + V²/2β  = d

=>V²/α  + V²/β = 2d

d = Area under graph =  (1/2)(T) * V

=> V = 2d/T

=> 4d²/T²α  + 4d²/T²β = 2d

=> 2d/α + 2d/β = T²

=> 1/α + 1/β = T²/2d

Answer 2

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