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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Updated on: 10-Oct-2022

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