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JEE Main & Advanced JEE Main Paper Phase-I (Held on 09-1-2020 Evening)

  • question_answer
    A particle starts from the origin at \[t=0\]with an initial velocity of \[3.0\,\hat{i}\,\,m\text{/}s\] and moves in the \[x\text{-}y\]plane with a constant acceleration\[\left( 6.0\,\hat{i}+4.0\hat{j} \right)m\text{/}{{s}^{2}}\]. The x-coordinate of the particle at the instant when its y-coordinate is \[32\text{ }m\]is D meters. The value of D is      [JEE MAIN Held on 09-01-2020 Evening]

    A) \[60\]

    B) \[32\]

    C) \[40\]

    D) \[50\]

    Correct Answer: A

    Solution :

    \[\frac{1}{2}\times 4\times {{t}^{2}}=32\] \[t=4\,s\] \[x=3\times 4+\frac{1}{2}\times 6\times {{4}^{2}}=60\]


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