A particle moving with a velocity of 2m/s.After covering 63 m it velocity is observe to be 16m/s.Calculate it acceleration.
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Answered by
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Answer:
initial velocity = 2 m/s
final Velocity = 16 m/s
distance covered = 63 m
using 3rd equation of motion that is 2as = (v)² - (u)²
so ,
2 × a × 63 = (16)² - (2)²
126 × a = 256 - 4
126 × a = 252
a = 252/126
so a = 2 m/s²
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