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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Aicharya · Accepted Answer

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 $x=\frac{1}{\beta} \cdot \log (\beta t+1)$

A particle is moving along $X$-axis with velocity $v=e^{-\beta x}$. At time $t=0$, the particle is located…

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