JEE Main & Advanced Physics NLM, Friction, Circular Motion JEE PYQ - NLM Friction Circular Motion

  • question_answer
    A point P moves in counter-clockwise direction on a circular path as shown in the figure. The movement of ‘P’ is such that it sweeps out a length\[s={{t}^{3}}+5,\]where s is in metres and t is in seconds. The radius of the path is 20 m. The acceleration of ‘P’ when\[t=2s\] is nearly. [AIEEE 2010]

    A) \[14\text{ }m/{{s}^{2}}\]

    B) \[13\text{ }m/{{s}^{2}}\]

    C) \[12\text{ }m/{{s}^{2}}\]

    D) \[7.2\text{ }m/{{s}^{2}}\]

    Correct Answer: C

    Solution :

    [a] \[s={{t}^{3}}+5\]
    \[a=\sqrt{(a_{r}^{2}+a_{t}^{2})}\]\[=\sqrt{{{\left( \frac{{{v}^{2}}}{R} \right)}^{2}}+{{\left( \frac{dv}{dt} \right)}^{2}}}=14\,m/{{s}^{2}}\]                       
    At \[t=2s\]

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