Uttarakhand PMT Uttarakhand PMT Solved Paper-2011

  • question_answer
    A particle moving with a velocity equal to 0.4 m/s is subjected to an acceleration of 0.15 \[m/{{s}^{2}}\]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)  m/s

    D)  between 0.7 and 0.1 m/s

    Correct Answer: B

    Solution :

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


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