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A Body start from origin and move along x-axis such that velocity at any instant is given by 4t^(3)-2t is in second and velocity in m/s Find acceleration of the particle when it is at a distance of 2 m from the origin: |
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Answer» `28 ms^(-2)` `implies int dx =int (4t^(3)-2t)dt` `x=(4t^(4))/(4)-(2t^(2))/(2)implies x=t^(2)(t^(2)-1)` when x=2 then 2 =`t^(4)-t^(2) implies t^(4)-t^(2)-2 =0` `implies t^(4)-2t^(2)+t^(2)-2=0` `t^(2)(t^(2)-2)+1(t^(2)-2)=0` `implies (t^(2)-2)(t^(2)+1)=0 implies t^(2)=2 implies t=sqrt(2)` Now `a=(dv)/(dt)=12t^(2)-2 implies a=12xx2-2=22m//s^(2)` |
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