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JEE Main & Advanced AIEEE Solved Paper-2010

  • 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  Solved  Paper-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: A

    Solution :

    \[s={{t}^{3}}+5\] \[v=\frac{ds}{dt}=3{{t}^{2}}\]                     \[\frac{dv}{dt}=6t\] \[a=\sqrt{(a_{r}^{2}+a_{t}^{2})}\]\[=\sqrt{{{\left( \frac{{{v}^{2}}}{R} \right)}^{2}}+{{\left( \frac{dv}{dt} \right)}^{2}}}=14\,m/{{s}^{2}}\]                At \[t=2s\]


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