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RAJASTHAN ­ PET Rajasthan PET Solved Paper-2009

  • question_answer
    A particle moves along with x-axis. The position\[x\]of particle with respect to time t from origin given by \[x={{b}_{0}}+{{b}_{1}}t+{{b}_{2}}{{t}_{2}}\]. The acceleration of particle is

    A)  \[{{b}_{0}}\]                

    B)  \[{{b}_{1}}\]

    C)  \[{{b}_{2}}\]                

    D)  \[2{{b}_{2}}\]

    Correct Answer: A

    Solution :

     Distance, \[x={{b}_{0}}+{{b}_{1}}t+{{b}_{2}}{{t}^{2}}\] Velocity \[v=\left( \frac{dx}{dt} \right)\] \[={{b}_{1}}+2{{b}_{2}}t\] Acceleration, \[\alpha =\frac{{{d}^{2}}x}{d{{t}^{2}}}=2{{b}_{2}}\]


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