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# A train accelerates from 36 km/h to 54 km/h in 10 sec. Calculate the distance travelled by train.

Initial velocity of the train $u=36\ km/h=36\times\frac{5}{18}=10\ m/s$

Final velocity of the train $v=54\ km/h=54\times\frac{5}{18}\ m/s=15\ m/s$

Time $t=10\ sec.$

Acceleration $a=\frac{v-u}{t}=\frac{15-10}{10}=\frac{5}{10}=0.5\ m/s^2$

Let $s$ be the distance traveled by train.

On using the equation of motion $v^2=u^2+2as$

Or $15^2=10^2+2\times0.5\times s$

Or $225=100+s$

Or $s=225-100$

Or $s=125\ m$

Therefore, the distance traveled by train is $125\ m$.

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