CEE Kerala Engineering CEE Kerala Engineering Solved Paper-2004

  • question_answer
    The velocity of a boat\[X\]relative to a boat Y is \[5\hat{i}-2\text{ }\hat{j}\]and that of\[Y\]relative to another boat Z is\[9\hat{i}+4\hat{j}\]where\[\hat{i}\]and\[\hat{j}\] are the velocity of k not per hour, east and north respectively. Then the velocity is:

    A)  \[\frac{\sqrt{2}}{10}\,knot/h\]                 

    B)  \[\frac{10}{\sqrt{2}}\,knot/h\]

    C)  \[10\sqrt{2}\,knot/h\]

    D)         \[2\sqrt{10}\,knot/h\]

    E)  \[10\,knot/h\]

    Correct Answer: C

    Solution :

    \[{{({{v}_{rel}})}_{x\to y}}=5\hat{i}-2\hat{j}\] and     \[{{({{v}_{rel}})}_{y\to z}}=9\hat{i}+4\hat{j}\] \[\therefore \]  \[{{({{v}_{rel}})}_{x\to y}}+{{({{v}_{rel}})}_{y\to z}}\]                 \[=(5\hat{i}-2\hat{j})+(9\hat{i}+4\hat{j})=14\hat{i}+2\hat{j}\] \[\Rightarrow \]               \[|\overrightarrow{v}|=|14\hat{i}+2\hat{j}|=\sqrt{{{14}^{2}}+{{2}^{2}}}\]                 \[=\sqrt{200}=10\sqrt{2}\,knot/h\]


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