NEET AIPMT SOLVED PAPER SCREENING 2005

  • 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:            

    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.


You need to login to perform this action.
You will be redirected in 3 sec spinner