Question

Two trains are moving in the opposite directions at speeds of 43 km/h and 51 km/h respec- tively. The time taken by the slower train to cross a man sit- ting in the faster train is 9 sec- onds. What is the length (in metre) of the slower train?

Answer 1

Explanation:

Velocity of first train- 43km /h

Velocity of second train = 51km/h

According to the question,

When the two heads of each train meets at a common point the total speed of slow train is equal to 43+51km/h=94mk/h=26.11m/s

therefore distance of slow train=speed ×time

=26.11×9

=234.99metres

Answer 2

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$\underline{\large\mathcal\red{solution}}$

As the both train are moving in opposite directions so the resultant velocity of the slower train will be =(43+51)=94km/hr

now ..94km/hr=(94000/3600)m/sec

now the slower train crossed a man standing on the faster train ...therefore actually it covers its length in 9 seconds.,

therefore ..lenght of the slower train was...

$\frac{94000}{3600} \times 9 \\ = \frac{94000}{400} \\ = 235 \: \: metres$

therefore the length of the slower train was=235 metres

$\underline{\large\mathcal\red{hope\: this \: helps \:you......}}$