Question

Shatabdi Express starting from rest attains is velocity of 108 km per hour in 2 minutes assuming the acceleration to be uniform find the value of acceleration the distance travelled by the train for attaining this velocity

Answer 1

Use equations of motion:
1. convert the Velocity to m/s: 108*(5/18)=30m/s
2. convert time to seconds: 120secs.
3. From the question, the initial velocity (u) is Zero and the final Velocity (v) is 30m/s in time(t)=120Secs
So from the equation of motion : v=u+a*t
therefore 'a' is 0.25m/s²
and S=u*t+0.5*a*t²
So S=0+0.5*120²
S=7,200m

Answer 2

Answer:Use equations of motion:

1. convert the Velocity to m/s: 108*(5/18)=30m/s

2. convert time to seconds: 120secs.

3. From the question, the initial velocity (u) is Zero and the final Velocity (v) is 30m/s in time(t)=120Secs

So from the equation of motion : v=u+a*t

therefore 'a' is 0.25m/s²

and S=u*t+0.5*a*t²

So S=0+0.5*120²

S=7,200m

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Explanation:this is the answer