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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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Answer» 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. Read more on Brainly.in - brainly.in/question/1252634#readmore |
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