Answer :

Given,

The initial velocity of both train A and train B, u = 72 km/h

"

Length of each train, l = 400 m

Acceleration of train A = 0 ms^{-2}

Acceleration of train B = 1 ms^{-2}

Time to overtake, t = 50 s

Distance to be covered to overtake,

Let, Distance traveled by train B __in "t"__ time= S_{B}

Distance traveled by train A in "t" time = SA

We have,

s_{B} + s_{A}= initial distance between the trains (d)

From 1^{st} equation of motion,

v = u + at

where,

v = Final velocity

u = Initial velocity

a = Acceleration/Deceleration

t = Time

Final velocity of train A, v_{A} = 20 + 0 = 20 m/s

Final velocity of train B, v_{B} = 20 + (1×50) = 70 m/s

From 2^{nd} equation of motion,

s = u × t + 0.5 × a × t^{2}

where,

u = Initial velocity

a = Acceleration or Deceleration

s = Distance covered

t = Time

∴

s_{A} = (20×50) + (0.5×0×50^{2}) = 1000 m

s_{B} = (20×50) + (0.5×1×50^{2}) = 2250 m

∴ Initial distance between trains, d = 2250 -1000

= 1250 m

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