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RAJASTHAN ­ PET Rajasthan PET Solved Paper-2011

  • question_answer
    The x and y coordinates of a particle at any time t are given\[x=7t+4{{t}^{2}}\]and\[y=5t,\]where\[x\]and y are in metre and t in second. The acceleration of particle at\[t=5s\]is

    A)  zero            

    B)  \[\text{8 }m/{{s}^{2}}\]

    C)  \[20\text{ }m/{{s}^{2}}\]         

    D)  \[\text{40 }m/{{s}^{2}}\]

    Correct Answer: B

    Solution :

     \[a={{\left[ \left( \frac{{{d}^{2}}x}{d{{t}^{2}}} \right)+\left( \frac{{{d}^{2}}y}{d{{t}^{2}}} \right) \right]}^{1/2}}\] Here         \[\frac{{{d}^{2}}y}{d{{t}^{2}}}=0\] Hence,    \[a=\frac{{{d}^{2}}x}{d{{t}^{2}}}=8\,m/{{s}^{2}}\]


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