Question

a train starting from rest and moving with uniform acceleration attains a speed of 90 kilometre per hour in 5 minutes find the acceleration and the distance traversed​

Answer 1

As, the train is at rest, Initial velocity U=0,

final velocity =90km/h=25m/s

t=5min=300s

we know that, V=U+at,

by putting given values,

25=0+a*300

a=25/300=.0833m/second square.

Answer 2

Answer:

The acceleration and distance traversed by the train are $1080km/h^{2}$ and 3.75km respectively.

Explanation:

Given that,

A train which starts from rest moves with uniform acceleration and attains a velocity of 90km/h in a time of 5 minutes.

We are required to find the acceleration and distance traversed by train.

As the train starts from rest,its initial velocity is $u=0km/h$ and

Time is $t=5min=\frac{5}{60}hr=\frac{1}{12} hr$

To obtain its acceleration we shall use the equation of motion which is mentioned below

$v=u+at$

Substituting the values,we get

$90=0+a(\frac{1}{12})= > a=1080km/h^{2}$

To find the distance we shall use the equation

$s=ut+\frac{1}{2}at^{2}$

Substituting the known values we get,

$s=0+\frac{1}{2}(1080)\frac{1}{12^{2}} =3.75km$

Therefore,

The acceleration and distance traversed by the train are $1080km/h^{2}$ and 3.75km respectively.

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