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 :
\[x=a{{e}^{-\alpha t}}+b{{e}^{\beta t}}\] Velocity \[v=\frac{dx}{dt}=\frac{d}{dt}(a{{e}^{-\alpha t}}+b{{e}^{\beta t}})\] \[=a.{{e}^{-\alpha t}}(-\alpha )+b{{e}^{\beta t}}.\beta )\] \[=-a\alpha {{e}^{-\alpha t}}+b\beta {{e}^{\beta t}}\] Acceleration \[=-a\alpha {{e}^{-\alpha t}}(-\alpha )+b\beta {{e}^{bt}}.\beta \] \[=a{{\alpha }^{2}}\,{{e}^{-\alpha t}}+b{{\beta }^{2}}{{e}^{\beta \,t}}\] Acceleration is positive so velocity goes on increasing with time.
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