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=  SB

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


We have,


sB + sA= initial distance between the trains (d) 

From 1st equation of motion,


v = u + at


where,


v = Final velocity


u = Initial velocity


a = Acceleration/Deceleration


t = Time


Final velocity of train A, vA = 20 + 0 = 20 m/s


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


From 2nd equation of motion,


s = u × t + 0.5  × a × t2


where,


u = Initial velocity


a = Acceleration or Deceleration


s = Distance covered


t = Time



sA = (20×50) + (0.5×0×502) = 1000 m


sB = (20×50) + (0.5×1×502) = 2250 m


Initial distance between trains, d = 2250 -1000


= 1250 m

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