A body starts from the origin and moves along the x -axis such that velocity at any instant is given by (4 t^3-2 t ) , where t is in second and velocity is in m / s . What is the acceleration of the particle, when it is 2 m from the origin?

Step-by-step solution

aligned & Given that v=4 t^3-2 t & x= v d t, x=t^4-t^2+C , at t=0, x=0 & C=0 aligned When particle is 2 m away from the origin, then ( aligned & x=t⁴-t² & as x=2 & t⁴-t²-2=0 aligned ) aligned & (t^2-2 ) (t^2+1 ) & =0 t & = 2 ~s aligned aligned & a= d v d t = d d t (4 t^3-2 t ) & a=12 t^2-2 & t= 2 sec & a=12 ( 2 )^2-2=22 ~m / s ^2 aligned