• # question_answer The displacement$x$of a particle moving along a straight line varies with time t as, $x=a{{e}^{-\alpha t}}+b{{e}^{-\beta t}},$where$a,b,\alpha$and$\beta$are positive constants. The velocity of the particle will A) go on decreasing with timeB) be independent of and $\beta$C) drop to zero when $\alpha =\beta$D) go on increasing with time

Solution :

$x=a{{e}^{-\alpha t}}+b{{e}^{\beta t}}$ $V=\frac{dx}{dt}=-a\alpha {{e}^{-\alpha t}}+b\beta {{e}^{\beta t}}$ Acceleration $=A=\frac{dv}{dt}=a{{\alpha }^{2}}{{e}^{-\alpha t}}+b{{\beta }^{2}}{{e}^{\beta t}}$ As acceleration is positive so the value of the velocity will go on Increasing Hence, the correction option is (d).

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