The displacement $x$ of a particle varies with time $t$ as $x=a e^{-a t}+B^{\beta x}$, where $a, b, \alpha$…
The displacement $x$ of a particle varies with time $t$ as $x=a e^{-a t}+B^{\beta x}$, where $a, b, \alpha$ and $\beta$ are positive constants. The velocity of the particle will:
be independent of $\beta$
drop to zero when $\alpha=\beta$
go on decreasing with time
go on increasing with time
Solution
From question
$\begin{aligned}
& x=a e^{-\alpha \mathrm{t}}+b e^{\beta t} \\
& V=\frac{d x}{d t}=-a \alpha e^{-x t}+b \beta_e^{\mathrm{B}+} \\
& \Rightarrow \mathrm{V}=-a \alpha e^{-t}+b \beta \varepsilon^{\beta t}
\end{aligned}$
$\therefore$ We can tell that velocity will increase with time since the negative component will decrease with an increase in time and the positive component will increase in the equation given equation.