CET Karnataka Engineering CET - Karnataka Engineering Solved Paper-2003

  • question_answer
    A student is standing at a distance of 50 metre from the bus. As soon as the bus begins its motion with an acceleration of \[1\text{ }m{{s}^{-2}}\], the student starts running towards the bus with a uniform velocity u. Assuming the motion to be along a straight road, the minimum value of n, so that the student is able to catch the bus is:

    A)  \[8\text{ }m{{s}^{-1}}\]                               

    B)  \[5\text{ }m{{s}^{-1}}\]

    C)  \[12\text{ }m{{s}^{-1}}\]             

    D)  \[10\text{ }m{{s}^{-1}}\]

    Correct Answer: D

    Solution :

    The student is able to catch the bus if in time t the distance travelled by him is equal to the distance travelled by bus in time t i.e.          \[{{s}_{1}}={{s}_{2}}\]                                                    ... (i) From eq. (1)                 \[0+\frac{1}{2}a{{t}^{2}}=ut-d\] or            \[{{t}^{2}}-2ut+2d=0\] It is quadratic equation So,          \[t=\frac{-2\pm \sqrt{4{{u}^{2}}-8ad}}{2}\]                 \[=\frac{-2\pm 2\sqrt{{{\mu }^{2}}-2ad)}}{2}\] For t to be real                 \[u\ge \sqrt{2ad}\ge \sqrt{2\times 1\times 50}=10\,m/s\]                 \[=10\,\,m{{s}^{-1}}\]


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