Question

Help!!

Two trains A and B of Length 400m each are moving on two parallel tracks with a uniform speed of 71km/h in the same direction with A ahead of B.The driver B decides to overtake A and accelerates by 1m/s .If after 50s the guard of B brushes past the driver A .what wad the original distance between them?​

Answer 1

$\large \rm {\underbrace{\underline{Elucidation:-}}}$

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$\sf \red {\underline{\underline{Provided\: that:}}}$

➻Length of two trains =400m

➻Uniform speed of Trains A and B=71km/h

➻We need to convert km/h into m/s

➻To perform such conversion,we need to multiply the value with the fraction,$\tt {\frac{5}{18}}$

➻Uniform speed of Trains A and B=$\tt {71×\frac{5}{18}}$

$\to \tt {U_{A}=U_{B}=19.72m/s}$

➻The driver B decides to overtake A and accelerates by 1m/s .If after 50s the guard of B brushes past the driver A

➻Acceleration(a) of train B =1m/s

➻As the train B overtakes train A and accelerates by 1m/s,the acceleration of train A is considered as zero

➻Acceleration(a) of train A =0m/s

➻However time time interval for both trains is 50s

➻Time interval(t)=50seconds

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$\sf \blue {\underline{\underline{To\: determine:}}}$

➻The original distance between train A and Train B.

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$\sf \orange {\underline{\underline{Distance\: covered\: by\: train\: A(s_{A})}}}$

$\to \tt {S_{A}=u_{A}×t}$

$\to \tt {S_{A}=19.72×50}$

$\implies \tt \green {\boxed{S_{A}=986m}}$

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$\sf \purple {\underline{\underline{Distance\: covered\: by\: train\: B(s_{B})}}}$

➻Using 2nd Kinematic equation:-

$\tt {\boxed{\underline{S=ut+\frac{1}{2}at^{2}}}}$

$\to \tt {S_{B}=u_{B}t+\frac{1}{2}a×t^{2}}$

$\to \tt {S_{B}=19.72×50+\frac{1}{2}×1×(50)^{2}}$

$\to \tt {S_{B}=986+\frac{1}{2}×2500}$

$\to \tt {S_{B}=986+\frac{1}{\cancel{2}}×{\cancel{2500}}^{1250}}$

$\to \tt {S_{B}=986+1250}$

$\implies \tt \green {\boxed{S_{B}=2236m}}$

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➻ Distance between the two trains,

$\implies \tt {S_{B}-S_{A}}$

$\implies \tt {2236-986}$

$\implies \tt \green {\fbox{\underline{1250m}}}$

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$\sf \pink {\underline{\underline{Thereupon,}}}$

➻The original distance between train A and Train B is 1,250m respectively.

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Answer 2

Answer:

Original distance between Train A and B is 1250 metres

Explanation:

Given,

Two trains A and B of length 400m each are moving on two parallel lines with a uniform speed of 71km/h

Here,

Train A is shead of Train B

And, A overtakes B within 50 seconds by accelerating 1m/s.

Find the original distance between A and B.

Original distance between both the trains is given by,

→ Distance covered by train A - Distance covered by train B.

Calculating distance covered by train A,

→ s = ut + ½.a.t²

Where,

• u(initial velocity) = 71km/h or, 19.72m/s(approx.)
• a(acceleration) = 1m/s
• t(time taken) = 50s

Then,

→ s = 19.72(50) + ½.(1)(50)²

→ s = 986 + 1250

→ s = 2236

Now, distance covered by B is :-

→ u*t

→ 19.72*50

→ 986

Hence, the original distance between Train A and B is :-

→ 2236 - 986

→ 1250

∴ Required answer: 1250 metres