मराठी

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 What Was the Original Distance Between Them - Physics

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प्रश्न

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?

उत्तर १

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_1 = ut + 1/2 a_1t^2`

= 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:

`s_u = ut + 1/2 at^2`

`=20xx50+1/2xx1xx(50)^2250 m`

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

shaalaa.com

उत्तर २

Here length of train A = length of train B = l = 400 m. As speed of both trains u = 72 km h-1 = 20 ms-1 in same direction, hence their relative velocity uBA = 0.

Let initial distance between the two trains be ‘S’ then train B covers the distance (S + 11) = (S + 800) m in time t = 50 s when accelerated with a uniform acceleration a = 1 m/s2.

`:. (s+800) = u_(AB) xx t + 1/2 at^2`

`= 0 + 1/2 xx1xx(50)^2 = 1250 m`

=> S = 1250 - 800 = 450 m

and initial distance between guard of train B from driver of train A = 450 + 800 = 1250 m.

shaalaa.com
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पाठ 3: Motion in a Straight Line - Exercises [पृष्ठ ५६]

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एनसीईआरटी Physics [English] Class 11
पाठ 3 Motion in a Straight Line
Exercises | Q 7 | पृष्ठ ५६

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