CMC Medical CMC-Medical Ludhiana Solved Paper-2014

  • question_answer
    A particle moving, with a velocity equal to 0.4 m/s is subjected to an acceleration of 0.15 \[m\text{/}{{s}^{2}}\] for 2 s in a direction at right angles to its direction of motion. The resultant velocity is

    A)  0.7 m/s

    B)  0.5 m/s

    C)  0.1 m/s

    D)  between 0.7 and 0.1 m/s

    Correct Answer: B

    Solution :

                    The will be no change in the velocity along horizontal direction so, \[{{v}_{x}}\]= 0.4 m/s. Velocity in perpendicular directions \[{{v}_{y}}=0+0.15\times 2=0.3\,\,m\text{/}s\] So, resultant velocity                 \[v=\sqrt{v_{x}^{2}+v_{y}^{2}}\]                 \[v=\sqrt{{{(0.4)}^{2}}+{{(0.3)}^{2}}}=0.5\,m\text{/}s\]


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