We are given a force that varies with time: $F\left(t\right)=F_0{e}^{-bt}$. Since the particle starts from rest at the origin, its initial velocity is zero. According to Newton's second law, the acceleration $a\left(t\right)$ is given by $a\left(t\right)=\frac{F\left(t\right)}{m}=\frac{F_0}{m}{e}^{-bt}$.

The velocity $v\left(t\right)$ is the integral of acceleration with respect to time. Let's compute it step by step.

Step 1: Set up the integral for velocity.
Since the particle starts from rest, $v\left(0\right)=0$. Therefore,
$v\left(t\right)={\int }_0^{t}a\left(s\right)ds={\int }_0^t\frac{F_0}{m}{e}^{-bs}ds$

Step 2: Evaluate the integral.
The integral of $e^{-bs}$ with respect to $s$ is $-\frac{1}{b}{e}^{-bs}$. So,
$v\left(t\right)=\frac{F_0}{m}[-\frac{1}{b}{e}^{-bs}{]}_0^t=\frac{F_0}{m}(-\frac{1}{b}{e}^{-bt}+\frac{1}{b})=\frac{F_0}{mb}(1-e^{-bt})$

Step 3: Analyze the behavior of $v\left(t\right)$.
The velocity function is $v\left(t\right)=\frac{F_0}{mb}(1-e^{-bt})$.

• At $t=0$, $e^{-b\times 0}=1$, so $v\left(0\right)=\frac{F_0}{mb}(1-1)=0$, which matches the initial condition.
• As $t\to \infty$, $e^{-bt}\to 0$, so $v\left(t\right)\to \frac{F_0}{mb}$. The velocity approaches a constant terminal value.
• The function $1-e^{-bt}$ increases from 0 and asymptotically approaches 1. Therefore, $v\left(t\right)$ starts at 0, increases, and approaches a constant value without decreasing or oscillating.

Step 4: Compare with the given curves.
The correct curve must start at the origin (0,0), increase, and approach a horizontal asymptote (a constant value) as time increases. It should not decrease, oscillate, or have any discontinuities.

Final Answer: The curve that depicts this behavior is the one that starts at the origin and increases, approaching a constant value asymptotically. Based on the typical representation of such functions, it is likely the first option (top-left image) if it shows a curve starting at (0,0) and leveling off.

Related Topics and Formulae

Newton's Second Law: $\overrightarrow{F}=m\overrightarrow{a}$ or $a=\frac{F}{m}$ for one-dimensional motion.

Velocity from Acceleration: $v\left(t\right)={\int }_{t_0}^{t}a\left(s\right)ds+v\left(t_0\right)$. Here, $t_0=0$ and $v\left(0\right)=0$.

Exponential Decay: The force $F\left(t\right)=F_0{e}^{-bt}$ decays exponentially from $F_0$ to 0. The parameter $b$ determines the rate of decay.