A train starting from rest attains a velocity of 72 km/h in 5 minutes. Assuming the acceleration is uniform, find the distance travelled by the train for attaining this velocity.

Here initial velocity $u=0$

Final velocity $v=72\ km/h=72\times\frac{5}{18}\ m/s$

$=20\ m/s$

Time $t=5\ minute=5\times60=300\ sec$

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

Or $a=\frac{20-0}{300}=\frac{1}{15}\ m/s^2$

Therefore, Distance travelled $s=ut+\frac{1}{2}at^2$

Or $s=0\times300+\frac{1}{2}\times\frac{1}{15}\times300^2$

Or $s=0+3000$

Or $s=3000\ m$

Or $s=3000\times\frac{1}{1000}\ km$

Or $s=3\ km$

Therefore, distance travelled, is $3\ km$.

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Updated on: 10-Oct-2022

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