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Haryana PMT Haryana PMT Solved Paper-2002

  • question_answer
    The position \[x\] of a particle varies with time \[t\,as\,x=a{{t}^{2}}-b{{t}^{3}}.\]The acceleration of the                 particle will be zero at time ( equal to :

    A) \[\frac{2a}{3b}\]                                              

    B)  \[\frac{1}{b}\]

    C) \[\frac{a}{3b}\]                                

    D)  \[0\]

    Correct Answer: C

    Solution :

                    \[x=a{{t}^{2}}-b{{t}^{3}}\] Velocity \[\upsilon =\frac{dx}{dt}=2at-3b{{t}^{2}}\] Acceleration \[a=\frac{{{a}^{2}}x}{dt}=2a-6bt\] substituting acceleration \[a=0\] \[\Rightarrow \]               \[2a-6\,bt=0\]                 \[t=\frac{2a}{6b}=\frac{a}{3b}\]


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