InterviewSolution
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InterviewSolution
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The position `x` of a particle with respect to time `t` along the x-axis is given by `x=9t^(2)-t^(3)` where `x` is in meter and `t` in second. What will be the position of this particle when it achieves maximum speed along the positive `x` directionA. 54mB. 81mC. 24mD. 32m |
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Answer» Correct Answer - a `x=9t^(2)-t^(3)` ...........(i) `therefore v=(dx)/(dt)=(d)/(dt)(9t^(2)-t^(3))=18-3t^(2)` For speed u to be maximum, the first derivative shou ld be zero and the second derivative shou ld be negative. `therefore (dv)/(dt)=18-6t` and `(d^(2)v)/(dt^(2))=-6` =18-6t t=3s Thus, the speed will be maximum at t=3s. From eqn. (i). `x=9(3)^(2)-(3)^(3)=54m` |
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