if the velocity of a train which starts from rest is 72 km/h after 5 minute,find out its acceleration and the distance travelled by the train in this time
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Explanation:
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Explanation:
u=0
v=72 km/h = 20m/s
t=5 min= 300 sec
from 1 eqn of motion -
v=u+at
20=0+a*300
20/300=a
a= 0.06 m/s^2
from 2 eqn of motion
v^2-u^2=2as
(20)^2-0^2=2*0.06*s
400=0.12s
S=400/0.12
= 3333.3m
another method
a=v/t
= 20/300
= 0.06m/s^2
v= s/t
s=vt
=20*300
6000m
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