Question

Two trains A and B of length 400 m
each are moving on two parallel tracks with
a uniform speed of 72 km h in the same
direction, with A ahead of B. The driver of
B decides to overtake A and accelerates
by 1 m s2. If after 50 s, the guard of B just
brushes past the driver of A, The original
distance between the two trains A, B is
1) 450 m
2) 1000 m
3) 1250 m 4) 1400 m​

Answer 1

the original distance between train a to b is 1250

Answer 2

Answer_:

$\longrightarrow{\large{\textsf{Original distance=450m}}}$

Step By Step Explaination_:

For train A :

• Initial velocity,u = 72 km/h = 20m/s
• Time, t= 50 s
• a= 0 (°•° uniform velocity)

From second eqⁿ of motion, distance (s1) covered by train A :

→ s1= ut+1/2at²

→ s1= 20×50+1/2×0×50²

→ s1= 20×50+0

→ s1= 1000m

For train B:

• u = 72 km/h = 20 m/s
• t= 50 s
• a= 1 m/s²

→s2= ut+1/2at²

→ 20×50+1/2×1×50²

→ 100+1/2×1×2500

→ s2= 2250 m

Hence the original distance between them :

Original distance= 1250- sum of the length of two trains

→[1250-(2×400)] m

→[1250-800] m

→ 450 m