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A train running at the speed of 60 km/hr crosses a pole in 9 seconds. What is the length of the train?

1120 metres
2180 metres
3324 metres
4150 metres

Answer:- 1

Explanation:-

Solution:

\begin{aligned} Speed = (60 \times \frac{5}{18}) m/sec = (\frac{50}{3}) m/sec \\ Length of the train = (Speed \times Time) \\ ∴ Length of the train \\ = (\frac{50}{3} \times 9) m = 150 m \end{aligned}

$\eqalign{ & {\text{Speed}} = \left( {60 \times \frac{5}{{18}}} \right){\text{m/sec}} = \left( {\frac{{50}}{3}} \right){\text{m/sec}} \cr & {\text{Length}}\,{\text{of}}\,{\text{the}}\,{\text{train}} = \left( {{\text{Speed}} \times {\text{Time}}} \right) \cr & \therefore {\text{Length}}\,{\text{of}}\,{\text{the}}\,{\text{train}} \cr & = \left( {\frac{{50}}{3} \times 9} \right)m = 150m \cr}$

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", "dateCreated": "7/24/2019 10:09:12 AM", "author": { "@type": "Person", "name": "Nitin Sir" } }, "suggestedAnswer": { "@type": "Answer", "text": "

Solution:

\begin{aligned} Speed = (60 \times \frac{5}{18}) m/sec = (\frac{50}{3}) m/sec \\ Length of the train = (Speed \times Time) \\ ∴ Length of the train \\ = (\frac{50}{3} \times 9) m = 150 m \end{aligned}

$\eqalign{ & {\text{Speed}} = \left( {60 \times \frac{5}{{18}}} \right){\text{m/sec}} = \left( {\frac{{50}}{3}} \right){\text{m/sec}} \cr & {\text{Length}}\,{\text{of}}\,{\text{the}}\,{\text{train}} = \left( {{\text{Speed}} \times {\text{Time}}} \right) \cr & \therefore {\text{Length}}\,{\text{of}}\,{\text{the}}\,{\text{train}} \cr & = \left( {\frac{{50}}{3} \times 9} \right)m = 150m \cr}$ ", "dateCreated": "7/24/2019 10:09:12 AM" } }

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