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A train starting from rest attains a velocity of $72\ km h^{-1}$ in $5\ minutes$. Assuming that the acceleration is uniform, find the distance travelled by the train for attainin this velocity.
Here, initial velocity $u=0$
Final velocity $v=72\ kmh^{-1}=72\times\frac{5}{18}=20\ ms^{-1}$
Time $t=5\ minutes=5\times60=300\ seconds$
Therefore, acceleration $a=\frac{v-u}{t}$
$=\frac{20-0}{300}$
$=\frac{1}{15}\ ms^{-2}$
On using the equation of motion $s=ut+\frac{1}{2}at^2$
$s=0\times300+\frac{1}{2}\times\frac{1}{15}\times300^2$
Or $s=0+3000$
Or $s=3000\ m$
Or $s=\frac{3000}{1000}\ km=3\ km$
Thus the distance traveled by train is $3\ km$.
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