A train starting from rest a trains a velocity of 72km/h in 5 minutes. Assuming the acceleration is uniform, find
1-the acceleration
2-the distance tavell by the train for attaining this velocity.
Answers
Answered by
1
u=0 m per sec
v= 72km/h=( 72×5)/18=20 m/sec
t = 5 min= 5×60=300 sec
1)acceleration=( v- u)/ t=(20-0)/300=2/30=0.067m/sec²
2)s= distance= ut +1/2at²=0×30 +((1/2)×a×t²)
=0+1/2 × 2/30×(300)²
=3000m
please vote me brainliest and follow me to get more answers easily..
v= 72km/h=( 72×5)/18=20 m/sec
t = 5 min= 5×60=300 sec
1)acceleration=( v- u)/ t=(20-0)/300=2/30=0.067m/sec²
2)s= distance= ut +1/2at²=0×30 +((1/2)×a×t²)
=0+1/2 × 2/30×(300)²
=3000m
please vote me brainliest and follow me to get more answers easily..
Similar questions