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Q. A body starts from the origin and moves along the axis such that the velocity at any instant is given by $ v=4t^{3}-2t $ where t is in second and the velocity is in $ m/s $ . Find the acceleration of the particle when it is at a distance of $ 2\,m $ from the origin.

Punjab PMETPunjab PMET 2010Motion in a Straight Line

Solution:

Velocity $ v=4t^{3}-2t $ ... (i)
$ \frac{dx}{dt}=4\,\,{{t}^{3}}-2t $
On integration, we get
$ x=2=t^{4}-t^{2}={{\alpha }^{2}}-\alpha $ ... (ii)
( $if\,\,t^{2}=\alpha $)
$ {{\alpha }^{2}}-\alpha -2=0 $
$ (\alpha -2)(\alpha +1)=0 $
$ \alpha =2 $
$ \alpha =-1 $ , which is not possible
$ {{t}^{2}}=\alpha =2 $ or $ t=\sqrt{2} $
Differentiating Eq. (i), w.r.t. $ t $
$ \frac{dv}{dt}=12{{t}^{2}}-2 $
$ \alpha =12\times 2-2=22\,ms^{-2} $