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AMU Medical AMU Solved Paper-2015

  • question_answer
    At a certain time a particle has a speed of 18 m/s in positive x-direction and 2.4 s later its speed was 30 m/s in the opposite direction. What is the magnitude of the average acceleration of the particle during the 2.4 s interval?

    A)  \[20\,m/{{s}^{2}}\]                        

    B)  \[10\,m/{{s}^{2}}\]

    C)  \[5\,m/{{s}^{2}}\]                          

    D)  \[2.5\,m/{{s}^{2}}\]

    Correct Answer: A

    Solution :

                    Average acceleration \[=\frac{change\text{ }in\text{ }veloclty}{Time\text{ }taken}\]                                 \[=\frac{-30\,m/s-18\,m/s}{2.4}\]                                 \[=20\,m/{{s}^{2}}\]


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