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Step-by-Step Solution
Step 1: Write down the given force and apply Newton’s second law
We are given a time-dependent force:
$$F(t) = F_0 e^{-bt}.$$
Since force is related to mass and acceleration by Newton's second law ($F = m \frac{dv}{dt}$), we have:
$$m \frac{dv}{dt} = F_0 e^{-bt}.$$
Step 2: Rearrange the differential equation
Divide both sides by $m$ to isolate $\frac{dv}{dt}$:
$$\frac{dv}{dt} = \frac{F_0}{m} e^{-bt}.$$
Step 3: Integrate to find the velocity
Integrate both sides from $t = 0$ to $t = t$ for the time variable, and from $v = 0$ to $v = v(t)$ for the velocity (the particle starts from rest, so $v(0) = 0$):
$$\int_{0}^{v} dv = \int_{0}^{t} \frac{F_0}{m} e^{-b t'} \, dt'.$$
(Carefully using a dummy variable $t'$ for integration.)
Step 4: Perform the integration
The left side simply becomes $v$, and the right side is:
$$v = \frac{F_0}{m} \int_{0}^{t} e^{-b t'} \, dt'
= \frac{F_0}{m} \left[ -\frac{1}{b} e^{-b t'} \right]_{0}^{t}.$$
Therefore,
$$v = \frac{F_0}{m} \left( -\frac{1}{b} e^{-b t} + \frac{1}{b} e^{0} \right)
= \frac{F_0}{mb} \left( 1 - e^{-b t} \right).$$
Step 5: Analyze the velocity function
The expression for the velocity as a function of $t$ is:
$$v(t) = \frac{F_0}{mb} \Bigl[ 1 - e^{-b t} \Bigr].$$
As $t \to \infty$, $e^{-b t} \to 0$, so the velocity approaches the constant value
$$v_{\max} = \frac{F_0}{mb}.$$
Thus, the curve for $v(t)$ starts from zero and asymptotically increases to a maximum value of $\frac{F_0}{mb}.$
Answer
The correct graph is the one that starts at $v = 0$ at $t = 0$ and asymptotically approaches $\frac{F_0}{mb}$ as $t \to \infty$.
Correct Answer Image:
