Two trains A and B of length 400 m each are moving on two parallel tracks with uniform speed of 72 kmh^-1

Question

Two trains A and B of length 400 m each are moving on two parallel tracks with a uniform speed of 72 kmh^-1 in the same direction with A ahead of B. The driver of B decides to overtake A and 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?

Answer 1

Train A 

Initial Velocity = U_A = 72 kmh^-1 = 20 ms^-1 

Time Taken = t = 50s 

Acceleration = a_A = 0 ms^-2 

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

\(s_A =u_At + \frac{1}{2}a_At^2\)

\(s_A =20\times50\)

\(s_A =1000\,m\)

Driver is at starting of Train A 

Train B 

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

Acceleration = a_B = 1 ms^-2 

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

\(s_B = u_B +\frac{1}{2}a_Bt^2\)

\(s_B =20\times50 +\frac{1}{2}\times1\times50^2\)

\(s_B = 2250\,m\)

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