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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.

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.