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(D) The velocity of the particle is given by $v = e^{-\beta x}$.Since $v = \frac{dx}{dt}$,we have $\frac{dx}{dt} = e^{-\beta x}$.Rearranging the terms

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$\frac{1}{\beta} \log [1+\beta t]$

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(D) The velocity of the particle is given by $v = e^{-\beta x}$.Since $v = \frac{dx}{dt}$,we have $\frac{dx}{dt} = e^{-\beta x}$.Rearranging the terms to separate the variables,we get $e^{\beta x} dx = dt$.Integrating both sides with the initial condition that at $t = 0$,$x = 0$:$\int_0^x e^{\beta x} dx = \int_0^t dt$$\left[ \frac{e^{\beta x}}{\beta} \right]_0^x = [t]_0^t$$\frac{1}{\beta} (e^{\beta x} - e^0) = t - 0$$\frac{1}{\beta} (e^{\beta x} - 1) = t$$e^{\beta x} - 1 = \beta t$$e^{\beta x} = 1 + \beta t$Taking the natural logarithm on both sides:$\beta x = \log(1 + \beta t)$$x = \frac{1}{\beta} \log(1 + \beta t)$

$A$ particle is moving along the $X$-axis with velocity $v = e^{-\beta x}$. At time $t = 0$,the particle is located at $x = 0$. The displacement of the particle as a function of time is

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