Question

Train moveing with velocity of 54kmh-1 is accelerated so that its velocity becomes 72kh-1 in 15 sec find the acceleration and distance coverd by train.
ans=

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Answer 1

Given:-

▪︎Initial velocity of the train(u)=54km/h

=>54×5/18

=>15m/s

▪︎Final velocity of the train(v)=72km/h

=>72×5/18

=>20m/s

▪︎Time(t)=5s

To find:-

•Acceleration(a)

•Distance covered by the train(s)

Solution:-

By,the 1st equation of motion,we know that----

=>v=u+at

=>a=v-u/t

=>a=20-15/15

=>a=5/15

=>a=1/3m/s² or 0.33m/s²

Now,by the 2nd equation of motion,we know that-----

=>s=ut+1/2at²

=>s=15×15+1/2×(1/3)×(15)²

=>s=225+37.5

=>s=262.5m

Thus-------

i.Acceleration is 1/3m/s² or 0.33m/s²

ii.Distance covered by the train is 262.5m.

Answer 2

Step-by-step explanation:

Given:-

▪︎Initial velocity of the train(u)=54km/h

=>54×5/18

=>15m/s

▪︎Final velocity of the train(v)=72km/h

=>72×5/18

=>20m/s

▪︎Time(t)=5s

To find:-

•Acceleration(a)

•Distance covered by the train(s)

Solution:-

By,the 1st equation of motion,we know that----

=>v=u+at

=>a=v-u/t

=>a=20-15/15

=>a=5/15

=>a=1/3m/s² or 0.33m/s²

Now,by the 2nd equation of motion,we know that-----

=>s=ut+1/2at²

=>s=15×15+1/2×(1/3)×(15)²

=>s=225+37.5

=>s=262.5m

Thus-------

i.Acceleration is 1/3m/s² or 0.33m/s²

ii.Distance covered by the train is 262.5m.

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