A) go on decreasing with time
B) be independent of and \[\beta \]
C) drop to zero when \[\alpha =\beta \]
D) go on increasing with time
Correct Answer: D
Solution :
\[x=a{{e}^{-\alpha t}}+b{{e}^{\beta t}}\] \[V=\frac{dx}{dt}=-a\alpha {{e}^{-\alpha t}}+b\beta {{e}^{\beta t}}\] 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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