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–1 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, what was the original distance between them with diagram and explanation plz?​

Answer 1

Answer:

Given that,

uA=uB=72kmh−1=72×185=20ms−1

Using the relations, s=ut+21at2, we get

SB=uBt+21at2=20×50+21×1×(50)2

SB=1000+1250=2250m

Also, let SA be the distance covered by the train A, then SA=uA×t

     =20×50=1000m

Original distance between the two trains

= SB−SA

=2250−1000=1250m

Answer 2

Required Answer

For train A:

Initial velocity, u = 72 km/h = 20 m/s

Time, t = 50 s

Acceleration, aI = 0 (Since it is moving with a uniform velocity)

From second equation of motion, distance (sI)covered by train A can be obtained as:

s = ut + (1/2)a1t2

= 20 × 50 + 0 = 1000 m

For train B:

Initial velocity, u = 72 km/h = 20 m/s

Acceleration, a = 1 m/s2

Time, t = 50 s

From second equation of motion, distance (sII) covered by train A can be obtained as:

sII = ut + (1/2)at2

= 20 X 50 + (1/2) × 1 × (50)2 = 2250 m

Length of both trains = 2 × 400 m = 800 m

Hence, the original distance between the driver of train A and the guard of train B is 2250 – 1000 – 800 = 450m.