a train starting from rest with the uniform acceleration of 0.2 meters per second for 5 minute calculate the speed and distance traveled with him
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Initial velocity (v)= 0m (as it starts from rest)
Acceleration (a)= 0.2m/s^2
Time (t)=5 minute
= 5×60 = 300s
Let final velocity be v
Using acceleration formula
a=(v-u)/t
0.2 = v/300
60 =v
So, final velocity was 60m/s.
Let distance be s
Using second equation of motion
s=ut+(1/2)at^2
s= (0)(300)+(1/2)(0.2)(300)^2
=9000m
So, the distance covered by the train was 9000m or 9km.
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Acceleration (a)= 0.2m/s^2
Time (t)=5 minute
= 5×60 = 300s
Let final velocity be v
Using acceleration formula
a=(v-u)/t
0.2 = v/300
60 =v
So, final velocity was 60m/s.
Let distance be s
Using second equation of motion
s=ut+(1/2)at^2
s= (0)(300)+(1/2)(0.2)(300)^2
=9000m
So, the distance covered by the train was 9000m or 9km.
Read more on Brainly.in - https://brainly.in/question/1252634#readmore
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