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. 24 mB. 32 mC. 54 mD. 81 m |
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Answer» Correct Answer - C The position x of a particle w.r.t. to time t along X-axis `x=9t^(2)-t^(3)` ….(i) Differentiating Eq. (i) w.r.t. time, we get speed, i.e., `upsilon = (dx)/(dt)=(d)/(dt)(9t^(2)-t^(3)) rArr upsilon = 18t-3t^(2)` …(ii) Again differntiating Eq. (ii) w.r.t. time, we get acceleration, i.e., `a=(d upsilon)/(dt)=(d)/(dt)(18t-3t^(2)) rArr a=18 -6t` ...(iii) Now, when speed of particle is maximum, its acceleration is zero. `a= 0 rArr 18-6t=0 rArr t=3s` Putting in Eq. (i), we obtain position of particle at that time `x=9(3)^(2)-(3)^(3)=9(9)-27=81-27=54m` |
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