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A train passes a pole in \[5{\text{ }}sec\] and a platform of \[50{\text{ }}m\] long in \[10{\text{ }}sec\].Find the length of the train.

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Hint: While applying formula for speed, the speed of the train remains the same while passing a pole or crossing a platform. If you wish to calculate speed, the formula is ,$\dfrac{d}{t}$ remains which becomes the formula for speed $t$, similarly for distance and time we can obtain the formulas from this.


Complete step by step solution:
Let the length of the train be meters.
Case 1: when train passes a pole
Speed of the train,$s = \dfrac{{distance\,(train\,\,length)}}{{time}}$
$ \Rightarrow s = \dfrac{x}{5}m/s$ \[........\left( 1 \right)\]
Case 2: When train crosses the platform,
Speed of the train $ = \dfrac{{distance\,(platform\,\,length)}}{{time}}$
Since, the distance is length of the train and length of the platform (because train will pass the platform with its length),
Speed of train $ = $\[\dfrac{{length{\text{ }}of{\text{ }}train{\text{ }} + {\text{ length of }}platform}}{{Time}}\]
$ \Rightarrow s = \dfrac{{x + 50}}{{10}}m/s$ \[.......\left( 2 \right)\]
From \[\left( 1 \right){\text{ }}and{\text{ }}\left( 2 \right),\]we will get
$\dfrac{x}{5} = \dfrac{{x + 50}}{{10}}$
$ \Rightarrow \dfrac{x}{1} = \dfrac{{x + 50}}{2}$
By cross multiplying the terms in the above expression, we will get
\[2x = x + 50\]
\[ \Rightarrow 2x - {\text{x}} = 50\]
$ \Rightarrow x = 50$
$ \Rightarrow $Length of train\[ = 50\]meters.

Additional Information: We can convert units of speed in the following ways:
$\left( i \right){\text{ }}x{\text{ }}km/hr = \left( {x \times \dfrac{5}{{18}}} \right)m/s.$
$\left( {ii} \right)\;x{\text{ }}m/s = \left( {x \times \dfrac{{18}}{5}} \right)km/hr.$

Note: Time taken by a train of length (\[{\text{l}}\]meters) to pass object of length (\[{\text{b}}\]meters) is the time taken by train to cover \[\left( {l + b} \right)\] meters.