A particle is moving along x-axis with its position (x) varying with time (t) as x= t^4+ t^2+ t+ . The ratio of its initial velocity to its initial acceleration, respectively, is:

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Correct answer: D. : 2

Position of particle, x= t^4+ t^2+ t+ Velocity v= d x d t =4 t^3+2 t+ Initial velocity =v(t=0)= Acceleration a= d v d t =12 t^2+2 Initial acceleration =a(t=0)=2 v(t=0) a(t=0) = 2

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