JEE Main & Advanced Sample Paper JEE Main - Mock Test - 6

  • question_answer
    Blocks A and C start from rest and move to the right with acceleration \[{{a}_{A}}=12t\,\,m/{{s}^{2}}\] and\[{{a}_{C}}=3\,\,m/{{s}^{2}}\]. Here t is in seconds. The time when block B again comes to rest is

    A) \[2s\]                            

    B) \[1s\]

    C) \[3/2s\] 

    D) \[1/2s\]

    Correct Answer: D

    Solution :

    [d] From constant relations, we can see that acceleration of block B is \[{{a}_{B}}=\left( \frac{{{a}_{A}}+{{a}_{C}}}{2} \right)\] with proper signs Hence, \[{{a}_{B}}=\left( \frac{3-12t}{2} \right)=1.5-6t\] or \[\frac{d{{v}_{B}}}{dt}=1.5-6t\] or  \[\int\limits_{0}^{{{v}_{B}}}{d{{v}_{B}}=\int\limits_{0}^{1}{(1.5-6t)dt}}\] or  \[{{v}_{B}}=\frac{1}{5}t-3{{t}^{2}}\] or \[{{v}_{B}}=0\] at \[t=\frac{1}{2}s\]

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