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JEE Main & Advanced Sample Paper JEE Main - Mock Test - 6

  • question_answer
    At \[t=0,\] a particle at \[(1,0,0)\] moves towards point \[(3,4,12)\] with a constant velocity of magnitude\[65\text{ }m/s\]. The position of the particle is measured in metres and time in seconds. Assuming constant velocity, the position of the particle at \[t=2\text{ }sec\]is

    A) \[\left( 13\hat{i}-120\hat{j}+40\hat{k} \right)m\]

    B) \[\left( 40\hat{i}+31\hat{j}-120\hat{k} \right)m\]

    C) \[\left( 13\hat{i}-40\hat{j}+12\hat{k} \right)m\]

    D) \[\left( 31\hat{i}+40\hat{j}+120\hat{k} \right)m\]

    Correct Answer: D

    Solution :

    [d] \[\vec{v}=\left( 65m/s \right)\frac{3\hat{i}+4\hat{j}+12\hat{k}}{\sqrt{{{3}^{2}}+{{4}^{2}}+{{(12)}^{2}}}}\] \[=65\times \frac{3\hat{i}+4\hat{j}+12\hat{k}}{13}=15\hat{i}+20\hat{j}+60\hat{k}\,\,m/s\] After \[2\text{ }sec,\] displacement \[\vec{S}=30\hat{i}+40\hat{j}+120\hat{k}\] Displacement = Change in positions \[={{\vec{P}}_{f}}-{{\vec{P}}_{s}}=\vec{S}=30\hat{i}+40\hat{j}+120\hat{k}\] Hence, final position \[={{\vec{P}}_{f}}=\left( 31\hat{i}+40\hat{j}+120\hat{k} \right)m\]


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