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# A train accelerated from $10\ km/hr$ to $40\ km/hr$ in $2\ minutes$. How much distance does it cover in this period?

Initial velocity of the train $u=10\ km/hr$

Final velocity of the train $v=40\ km/hr$

Time taken in changing the velocity $t=2\ minutes$

$=\frac{2}{60}\ hour$

$=\frac{1}{30}\ hour$

Therefore, acceleration $a=\frac{v-u}{t}$

$=\frac{40-10}{\frac{1}{30}}=30\times30=900\ km/hour^2$

Using the equation, $s=ut+\frac{1}{2}at^2$

$=10\times\frac{1}{30}+\frac{1}{2}\times900\times( \frac{1}{30})^2$

$=\frac{1}{3}+\frac{1}{2}\times900\times\frac{1}{900}$

$=\frac{1}{3}+\frac{1}{2}$

$=\frac{2+3}{6}$

$=\frac{5}{6}\ km$

$=0.833\ km$

$=0.833\times1000\ km$

$=833\ m$

Thus, the distance covered by the train during $2\ minutes$ is $833\ meter$.

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