A train starting from rest attains a velocity of 72 kmh-1 in 5 minute assuming that the acceleration is unuform find the acceleration and distance travelled by the train for attaining this velocity
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v=u+at
72*1000/3600=o+a*5*60
20=300a
a=20/300
a=1/15m/sec^2
s=ut+1/2at^2
s=1/2*1/15*300*300
s=1/30*300*300
s=1000m
1km
72*1000/3600=o+a*5*60
20=300a
a=20/300
a=1/15m/sec^2
s=ut+1/2at^2
s=1/2*1/15*300*300
s=1/30*300*300
s=1000m
1km
simran4137:
your answer is wrong
1/30×300×300=3000m=3km
ohh yes its 3km...typo error
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