A particle is moving along the path given by `y=(C )/(6)t^(6)` (where `C` is a positive constant). The relation between the acceleration (`a`) and the velocity (`v`) of the particle at `t=5sec` isA. 5a = vB. a = 5vC. `a = sqrt(v)`D. a = v

A particle is moving along the path given by `y=(C )/(6)t^(6)` (where

1.	A particle is moving along the path given by `y=(C )/(6)t^(6)` (where `C` is a positive constant). The relation between the acceleration (`a`) and the velocity (`v`) of the particle at `t=5sec` isA. 5a = vB. a = 5vC. `a = sqrt(v)`D. a = v
Answer» Correct Answer - A Angular impulse (about com) = `J (L)/(2) = (MV(L)/(2))=(l_(cm))omega=((2ML^(2))/(4))omega`

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A particle is moving along the path given by `y=(C )/(6)t^(6)` (where `C` is a positive constant). The relation between the acceleration (`a`) and the velocity (`v`) of the particle at `t=5sec` isA. 5a = vB. a = 5vC. `a = sqrt(v)`D. a = v

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