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A body starts from the origin and moves along the X-axis such that the velocity at any instant is given by `(4t^3-2t)`, where t is in sec and velocity in m/s. what is the acceleration of the particle when it is 2 m from the origin?A. `28 ms^(-2)`B. `22 ms^(-2)`C. `12 ms^(-2)`D. `10ms^(-2)` |
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Answer» Correct Answer - 2 `v_(t)=4t^(3)-2t, " " (dx)/(dt)=4t^(3)-2t`, `int_(0)^(x) dx=int_(0)^(t) (4t^(3)-2t) dt` `x=t^(4)-t^(2)" " "When" x=2,t=?` `2=(t^(4)-t^(2))` `t^(4)-t^(2)-2=0, " " t^(2)=y` ltbRgt `y^(2)-y-2=0` `y=(1+pmsqrt(1-4xx1xx-2))/2=(1+pmsqrt(9))/2=(1+3)/2` `(1-3)/2` `y=2,-1" " t^(2)=2, -1` `t^(2)=-1` (impossible) `t=pmsqrt(2)` (possible), `a=(dv)/(dt)=12t^(2)-2` `a=24-2=22 m//s^(2)` |
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