Question

Two trains A and B of length 400m each are moving on two parallel tracks with uniform speed of 72 km per hour in the same direction with A ahead of B. the driver of B decides to overtake A and accelerates by 1 m per second. if after 50 seconds. the guard of B passes the driver of A what was the original distance between the two trains

Answer 1

Hii dear,

# Answer- 450m

# Explaination-
For train A:
Initial velocity, u = 72 km/h = 20 m/s
Time, t = 50 s
Acceleration, a1 = 0
From second equation of motion, distance (s1)covered by train A can be obtained as:
s1 = ut + (1/2)a1t^2
s1 = 20 × 50 + 0
s1 = 1000 m

For train B:
Initial velocity, u = 72 km/h = 20 m/s
Acceleration, a2 = 1 m/s2
Time, t = 50 s
From second equation of motion, distance (s2) covered by train A can be obtained as:
s2 = ut + (1/2)a2t^2
s2 = 20 X 50 + (1/2) × 1 × (50)2
s2 = 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.

Hope this is helpful.

Answer 2

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For train A:

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

Time, t = 50 s

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

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

S = ut + 1/2 at^2

= 20 × 50 + 0 = 1000 m

For train B:

Initial velocity,

u = 72 km/h = 20 m/s

Acceleration, a = 1 m/s^2

Time, t = 50 s

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

S = ut + 1/2 at^2

= 20× 50 + 1 × (50)^2 = 2250 m

Hence, the original distance between the driver of train A and the guard of train B = 2250 – 1000 = 1250 m.

I hope, this will help you

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