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The displacement x of a particle varies with time 't' as $x=ae^{-\alpha t}+b e^{\beta t}$ where $a,b,\alpha, \beta$ are positive constants. The velocity of the particle will

$\begin{array}{1 1} (a) go\;on\;decreasing \;with \;time \\(b)be\; independent\; of\; \alpha\;and\; \beta \\(c) drop\;of\;zero\;when\;\alpha=\beta \\(d)\;go\;on\;increasing\;with\;time \end{array} $

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Answer: go on increasing with time.
Given $x=ae^{-\alpha t}+be^{\beta t}$
Velocity = displacement / time = $v=\large\frac{dx}{dt}$$=-a\; \alpha e^{-\alpha t}+b \beta e^{\beta t}$
As you can see, velocity is dependent of $\alpha$ and $\beta$, and does not drop to zero when $\alpha = \beta$
Now as t increases, $-a\;\alpha e^{-\alpha}$ decreases and $b\;\beta e^{\beta}$ increases $\rightarrow$ velocity goes on increasing with time.
answered Jun 23, 2013 by meena.p
edited Aug 11, 2014 by balaji.thirumalai

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