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A particle is moving along $X$-axis with velocity $v=e^{-\beta x}$. At time $t=0$, the particle is located at $x=0$. The displacement of the particle as function of time is
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The correct answer is:
$\frac{1}{\beta} \log [1+\beta t]$
Velocity of particle,
$v=e^{-\beta x}$
$\Rightarrow \quad \frac{d x}{d t}=e^{-\beta x}$
or $\quad \frac{d x}{e^{-\beta x}}=d t$
or $\quad e^{\beta x} d x=d t$
$\Rightarrow$ Integrating, we get
$\int_0^x e^{\beta x} d x=\int_0^t d t$
$\begin{aligned} & \Rightarrow \quad \frac{1}{\beta}\left[e^{\beta x}\right]_0^x=[t]_0^t \\ & \Rightarrow \quad \frac{1}{\beta}\left(e^{\beta x}-e^0\right)=t-0 \Rightarrow \frac{1}{\beta}\left(e^{\beta x}-1\right)=t\end{aligned}$
$e^{\beta x}=\beta t+1$
Taking log, we get
$\log \left(e^{\beta x}\right)=\log (\beta t+1)$
$\begin{array}{ll}\Rightarrow & \beta x=\log (\beta t+1) \\ \Rightarrow & x=\frac{1}{\beta} \cdot \log (\beta t+1)\end{array}$
So, displacement function for the particle is
$x=\frac{1}{\beta} \cdot \log (\beta t+1)$
$v=e^{-\beta x}$
$\Rightarrow \quad \frac{d x}{d t}=e^{-\beta x}$
or $\quad \frac{d x}{e^{-\beta x}}=d t$
or $\quad e^{\beta x} d x=d t$
$\Rightarrow$ Integrating, we get
$\int_0^x e^{\beta x} d x=\int_0^t d t$
$\begin{aligned} & \Rightarrow \quad \frac{1}{\beta}\left[e^{\beta x}\right]_0^x=[t]_0^t \\ & \Rightarrow \quad \frac{1}{\beta}\left(e^{\beta x}-e^0\right)=t-0 \Rightarrow \frac{1}{\beta}\left(e^{\beta x}-1\right)=t\end{aligned}$
$e^{\beta x}=\beta t+1$
Taking log, we get
$\log \left(e^{\beta x}\right)=\log (\beta t+1)$
$\begin{array}{ll}\Rightarrow & \beta x=\log (\beta t+1) \\ \Rightarrow & x=\frac{1}{\beta} \cdot \log (\beta t+1)\end{array}$
So, displacement function for the particle is
$x=\frac{1}{\beta} \cdot \log (\beta t+1)$
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