a train starting from rest attains a velocity of 72 km/hr in 5min.calculate:- acceleration and distance travelled by train for attaining this velocity.
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Answered by
3
V=72km/h=20m/s
U=0
T=5min=5×60=300s
V=u+at
a=v-u/t
=20/300=1/15m/s^2
S=ut+1/2at^2
=0×300+1/2×1/15×300×300
Calculation do yrself
U=0
T=5min=5×60=300s
V=u+at
a=v-u/t
=20/300=1/15m/s^2
S=ut+1/2at^2
=0×300+1/2×1/15×300×300
Calculation do yrself
Answered by
1
v=20m/s here ,
v=u +at
20=0+a300
20÷300=a
a=0.06m/s^2
and v^2=u^2 +2as
400=0+2*0.06s
400=0.12s
400÷0.12=s
3333m =s
you can also take acceleration 0.07m/s^2
then your distance
will=2857.1m
v=u +at
20=0+a300
20÷300=a
a=0.06m/s^2
and v^2=u^2 +2as
400=0+2*0.06s
400=0.12s
400÷0.12=s
3333m =s
you can also take acceleration 0.07m/s^2
then your distance
will=2857.1m
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