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` is

Options

A. 5a = v
B. a = 5v
C. `a = sqrt(v)`
D. a = v

Answer

A

Explanation

Angular impulse (about com) = `J (L)/(2) = (MV(L)/(2))=(l_(cm))omega=((2ML^(2))/(4))omega`