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-I in the same direction, with A ahead of B. The driver of
B decides to overtake A and accelerates by 1 m s-2. If after 50 s, the guard of B just
brushes past the driver of A, what was the original distance between them ?
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Answer 1

Answer:

850 m

Explanation:

Train A:-

Speed = 72km/h = 72×5/18 = 20 m/s

Time = 50 seconds

Distance = Speed × Time

= 20 × 50

= 1000 m

Train B:-

u = 72 km/h = 20 m/s

t = 50 seconds

a = 1 m/s²

s = ut + 1/2at²

= 20×50 + 1/2×1×(50)²

= 1000+ 1/2×2500

= 1000+1250

= 2250 m

more distance traveled by train B

= 2250-1000

= 1250 m

Length of 1 train = 400 m

Distance between them

= 1250-400

= 850 m

hope this helps u

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.