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$\textrm{The speed of train is given in km/hr & length of the train is asking in the meter.}$

$\textrm{To convert the speed of the train to km/hr to m/sec we have to multiply by 5/18.}$

$\because$ $\textrm{1km=1000m & 1hr=3600sec}$

$\therefore$ $\textrm{60km/hr= ($\frac{60*5}{18}$)m/sec}$

$\textrm{Speec Of train}$ $\Rightarrow$ ($\frac{50}{3})m/sec$

$Time=9sec$

$\because$ $\textrm{Speed=$\frac{Distance}{Time}$}$ $\Rightarrow$ $\textrm{Distance=Speed*Time}$

$\therefore$ $\textrm{Distance=($\frac{50}{3}*9$)m/sec}$

$\textrm{Distance=150m}$

$\therefore$ $\textrm{Length of train is150m,Option D.}$

$\textrm{To convert the speed of the train to km/hr to m/sec we have to multiply by 5/18.}$

$\because$ $\textrm{1km=1000m & 1hr=3600sec}$

$\therefore$ $\textrm{60km/hr= ($\frac{60*5}{18}$)m/sec}$

$\textrm{Speec Of train}$ $\Rightarrow$ ($\frac{50}{3})m/sec$

$Time=9sec$

$\because$ $\textrm{Speed=$\frac{Distance}{Time}$}$ $\Rightarrow$ $\textrm{Distance=Speed*Time}$

$\therefore$ $\textrm{Distance=($\frac{50}{3}*9$)m/sec}$

$\textrm{Distance=150m}$

$\therefore$ $\textrm{Length of train is150m,Option D.}$

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Given that, speed of the train $S = 60\;km/hr = 60 \times \dfrac{1000}{60 \times 60}\;m/sec = \dfrac{50}{3}\;m/sec$

and, time to cross the pole $T =9\;sec $

Let’s lenth of the train $ = D\;m$

We know that $S = \dfrac{D}{T}$

$\implies D = S \times T$

$\implies D = \dfrac{50}{3}\;m/sec \times 9\;sec$

$\implies D = 150\;m.$

So, the correct answer is $(D).$

and, time to cross the pole $T =9\;sec $

Let’s lenth of the train $ = D\;m$

We know that $S = \dfrac{D}{T}$

$\implies D = S \times T$

$\implies D = \dfrac{50}{3}\;m/sec \times 9\;sec$

$\implies D = 150\;m.$

So, the correct answer is $(D).$