Question

a train 100m long is running at the speed of 21 kilometre per hour and another train 150 metre long is running at the speed of 36 km per hour in the same direction how long will the faster train take to pass the first train ​

Answer 1

$\red{\underline{\underline{Answer:}}}$

$\sf{Faster \ train \ will \ take \ 61.2 \ seconds}$

$\sf{to \ cross \ first \ train.}$

$\sf\orange{Given:}$

$\sf{For \ first \ train,}$

$\sf{\implies{Length=100 \ m}}$

$\sf{\implies{Speed=21 \ km \ h^{-1}}}$

$\sf{For \ second \ train,}$

$\sf{\implies{Length=250 \ m}}$

$\sf{\implies{Speed=36 \ km \ h^{-1}}}$

$\sf\pink{To \ find:}$

$\sf{Time \ taken \ by \ second \ train}$

$\sf{to \ pass \ first \ train.}$

$\sf\green{\underline{\underline{Solution:}}}$

$\sf{Speed \ of \ second \ train \ - \ Speed}$

$\sf{of \ first \ train.}$

$\sf{\implies{36-21}}$

$\sf{\implies{15 \ km \ h^{-1}}}$

$\sf{If \ we \ consider \ speed \ of \ second \ train}$

$\sf{15 \ km \ h^{-1}. \ The \ speed \ of \ first}$

$\sf{train \ will \ be \ zero \ because }$

$\sf{both \ speed \ will \ be \ at \ same \ position}$

$\sf{at \ speed \ of \ 21 \ km \ h^{-1},}$

$\sf{but \ speed \ of \ second \ train \ is }$

$\sf{15 \ km \ hr^{-1} \ more.}$

$\sf{Hence, \ for \ speed \ 15 \ km \ h^{-1}}$

$\sf{of \ second \ train \ speed \ of \ first}$

$\sf{train \ will \ be \ zero.}$

$\sf{1 \ km=1000 \ m}$

$\sf{\therefore{15 \ km=15000 \ m}}$

$\sf{\implies{Speed=15000 \ m \ h^{-1}}}$

$\sf{Distance=Addition \ of \ length \ of \ trains}$

$\sf{\therefore{Distance=100+150}}$

$\sf{\therefore{Distance=250 \ m}}$

$\sf{Time(t)=\frac{Distance}{Speed}}$

$\sf{Time=\frac{250}{15000}}$

$\sf{\therefore{Time=0.017 \ hr (approx)}}$

$\sf{1 \ hr=3600 \ seconds}$

$\sf{\therefore{Time=0.017\times3600}}$

$\sf{\therefore{Time=61.2 \ seconds}}$

$\sf\purple{\tt{\therefore{Faster \ train \ will \ take \ 61.2 \ seconds}}}$

$\sf\purple{\tt{to \ cross \ first \ train.}}$

Answer 2

★$\color{darkblue}\underline{\underline{\sf Answer:}}$

$\rm\orange{Time \ taken \ by \ first \ train \ to \ cross }$

$\rm\orange{second \ train \ = 61.2 \ seconds}$

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$\sf\large\underline\red{Given:}$

$\rm{According \ to \ condition \ for \ first \ train,}$

⇒$\rm{Length \ = \ 100m}$

⇒$\rm{Speed \ = \ 21 \ km \ h^-1}$

$\rm{According \ to \ condition \ for \ second \ train,}$

⇒$\rm{Length \ = \ 250m}$

⇒$\rm{Speed \ = \ 36 \ km \ h^-1}$

__________________________________________

$\sf\large\underline\purple{To \ Find :}$

$\rm{Time \ taken \ by \ first \ train \ to \ cross }$

$\rm{second \ train}$

__________________________________________

★$\color{red}\underline{\underline{\sf Solution}}$

$\rm{Difference \ between \ the \ speeds \ of \ both }$

$\rm{trains}$

$\rm{Speed \ of \ first \ train \ speed \ of \ second \ train}$

$\rm{= \ 36 \ - \ 21}$

$\rm{= \ 15 \ km \ {hr}^{-1}}$

↦$\rm{If \ we \ consider \ speed \ of \ second \ train }$

$\rm{15 \ km \ {hr}^{-1}}$

↦$\rm{The \ speed \ of \ first \ train \ will \ be \ zero}$

$\rm{because \ both \ trains \ will \ be \ at \ same \ position}$

$\rm{When \ speed \ is \ 21 \ km \ {hr}{-1}}$

$\rm{but \ speed \ of \ second \ train \ 15 \ km \ {hr}{-1}}$

$\rm{more \ than \ that \ of \ first \ train}$

$\rm{we \ know,}$

$\rm{1 \ km \ = \ 1000 \ m}$

⇒$\rm{Speed \ = \ 15000 \ m \ {hr}{-1}}$

⇒$\rm{Distance \ = }$

$\rm{Length \ of \ first \ train \ + \ Length \ of \ second \ train}$

⇒$\rm{Distance \ = \ 100 \ + \ 150}$

⇒$\rm{Distance \ = \ 250 \ m}$

$\rm{Time \ = \ \frac{Distance}{Speed}}$

⇒$\rm{Time \ = \ \frac{250}{15000}}$

⇒$\rm{Time \ = \ 0.017 \ ( Approximate \ value )}$

$\rm{Now \ convert \ hour \ into \ seconds}$

$\rm{As \ we \ know,}$

⇒$\rm{1 \ hr \ = \ 3600 \ sec }$

⇒$\rm{Time \ = \ 0.017 \times \ 3600}$

⇒$\rm{Time \ = \ 61.2 \ sec }$

$\rm\orange{Time \ taken \ by \ first \ train \ to \ cross \ second }$

$\rm\orange{train \ = 61.2 \ seconds}$

__________________________________________

$\rm\green{Hope \ it \ helps \ :))}$