# Two trains A and B of length 400 m each are moving on two parallel tracks with uniform speed of 72 kmh-1 in the same direction with A ahead of B. The driver of B decides to overtake Aand accelerate by 1 ms-2. If after 50 s, the guard of B just passes the driver of A, what was the original distance between them?

Train A

Initial Velocity = UA = 72 kmh-1 = 20 ms-1

Time Taken = t = 50s

Acceleration = aA = 0 ms-2

Now, from the equations of motion, we know that,   m

Driver is at starting of Train A

Train B

Initial Velocity = UA = 72 kmh-1 = 20 ms-1 Time Taken = t = 50 s

Acceleration = aB = 1 ms-2

Now, from the equations of motion, we know that,   m

Guard is at the end of Train B

Now, length of both trains = 400 m + 400 m = 800 m

Now, Original distance between Train A and B is S, which can be obtained as given below:

S = 2250 m – 1000 m – 800 m

= 450 m

Rate this question :

How useful is this solution?
We strive to provide quality solutions. Please rate us to serve you better.
Related Videos  About Acceleration50 mins  Velocity Time Graph46 mins  Champ Quiz | Motion under gravity46 mins  Second Equation of Motion48 mins  Understanding Motion60 mins  Interactive Quiz - Motion29 mins  Learn The Use of Equations of Motion37 mins  Learning about Velocity54 mins  Foundation | Circular Motion51 mins  Quiz | Graph Part- 246 mins
Try our Mini CourseMaster Important Topics in 7 DaysLearn from IITians, NITians, Doctors & Academic Experts
Dedicated counsellor for each student
24X7 Doubt Resolution
Daily Report Card
Detailed Performance Evaluation view all courses 