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NEET AIPMT SOLVED PAPER SCREENING 2007

  • question_answer
                    The position x of a particle with respect to time t along x-axis is given by \[x=9{{t}^{2}}-{{t}^{3}}\] where x is in metre and t in second. What will be the position of this particle when it achieves maximum speed along the +x direction?                                                                                                                                                     

    A)                  32 m           

    B)                  54 m    

    C)                  81 m    

    D)                  24 m

    Correct Answer: B

    Solution :

                    Key Idea: At the instant when speed is maximum, its acceleration is zero.                 Given, die position \[x\] of a particle with respect to time \[t\] along \[x\]-axis                 \[x=9{{t}^{2}}-{{t}^{3}}\]                                              ...(i)                 Differentiating Eq. (i), with respect to time, we get speed, i.e.,                 \[v=\frac{dx}{dt}=\frac{d}{dt}(9\,{{t}^{2}}-{{t}^{3}})\]                 or            \[v=18\,t-3{{t}^{2}}\]                                     ...(ii)                 Again differentiating Eq. (ii), with respect to time, we get acceleration, i.e.,                 \[a=\frac{dv}{dt}=\frac{d}{dt}(18\,t-3{{t}^{2}})\]                 or         a = 18 - 6t                               ...(iii)                 Now, when speed of particle is maximum, its acceleration is zero, i.e.,                 a = 0                 i.e.,          18 ? 6t = 0 or t = 3s                 Putting in Eq. (i), we obtain position of particle at that time                 \[x=9{{(3)}^{3}}-{{(3)}^{3}}=9(9)-27\]                 \[=81-27=54\,m\]


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