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 :
We are given \[x=a{{e}^{-\alpha t}}+b{{e}^{\beta t}}\] Velocity \[v=\frac{dx}{dt}=\frac{d}{dt}\,\,(a{{e}^{-at}}+b{{e}^{\beta t}})\] \[=\,\,a\,\,.\,\,{{e}^{-\alpha \,t}}\,(-\alpha )+b{{e}^{\beta t}}.\,\,\beta \] \[=-a\alpha {{e}^{-\alpha t}}\,+\,\,b\beta {{e}^{\beta t}}\] Acceleration \[=-a\alpha {{e}^{-\alpha t}}\,\,(-\alpha )+b\beta {{e}^{\beta t}}.\,\beta \] \[=\,\,a{{\alpha }^{2}}{{e}^{-\alpha t}}+b{{\beta }^{2}}{{e}^{\beta t}}\] Acceleration is positive so velocity goes on increasing with time.
You need to login to perform this action.
You will be redirected in
3 sec
