a train starting from rest attains a velocity of 72 km/h in 5 minutes assuming the acceleration is uniform find the acceleration the distance travelled by the train for attaining this velocity.
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distance =(v^2-u^2/2a)
v=72*5/18=20m/s
u=0m/s
t=300sec
acc=(v-u)/t
acc=20/300=1/15m/s
distance=((20)^2-0^2)/2*1/15=400/(2/15)=400*(15/2)=3000 m
v=72*5/18=20m/s
u=0m/s
t=300sec
acc=(v-u)/t
acc=20/300=1/15m/s
distance=((20)^2-0^2)/2*1/15=400/(2/15)=400*(15/2)=3000 m
jhanvikalra31:
you are from Rich Harvest Public School
naa man
me from DAV
ok
thanks for answer
and you
i am from rich harvest public school
ooh nice
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