NEET Physics Motion in a Straight Line / सरल रेखा में गति NEET PYQ-One Dimensional Motion

  • question_answer
    The displacement \[x\] of a particle varies with time \[t\] as \[x=a{{e}^{-at}}+b{{e}^{\beta t}},\] where \[a,\,b,\,\,\alpha \] and \[\beta \] are positive constants. The velocity of the particle will:                               [AIPMT (S) 2005]

    A) go on decreasing with time                      

    B) be independent of \[\alpha \] and \[\beta \]

    C) drop to zero when \[\alpha =\beta \]                     

    D) go on increasing with time

    Correct Answer: D

    Solution :

    Given,   \[x=a{{e}^{-\alpha t}}+b{{e}^{\beta t}}\]
    So, velocity \[v=\frac{dx}{dt}\]
    \[=-a\alpha {{e}^{-\alpha t}}+b\beta {{e}^{\beta t}}\]
    \[=A+B\]
    where,   \[A=-a\alpha {{e}^{-}},\,B=b\beta {{e}^{\beta t}}\]
    The value of term \[A=-a\alpha {{e}^{-at}}\] decreases and of term \[B=b\beta {{e}^{\beta t}}\] increases with increase in time. As a result, velocity goes on increasing with time.


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