UPSC Physics Units and Measurements / मात्रक और मापन Question Bank Measurements and Motion

  • question_answer
    The displacement of a particle at time t is given by \[\vec{x}=a\hat{i}+bt\hat{j}+\frac{c}{2}{{t}^{2}}\hat{k}\] where a, b and c are positive constants. Then the particle is [NDA]

    A) accelerated along \[\frac{18}{5}\] direction

    B) decelerated along \[\frac{18}{5}\] direction

    C) decelerated along \[\hat{j}\] direction

    D) accelerated along \[\hat{j}\] direction

    Correct Answer: A


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