A particle of mass `m` is at rest the origin at time ` t= 0 ` . It is subjected to a force ` F(t) = F_(0) e^(-bt)` in the `x` - direction. Its speed ` v(t)` is depicted by which of the following curves ?

Options

A.
B.
C.
D.

Explanation

To solve the problem, we need to determine the velocity \( v(t) \) of a particle subjected to a force \( F(t) = F_0 e^{-bt} \) in the x-direction. Here are the steps to derive the expression for velocity:

### Step 1: Write down the relationship between force, mass, and acceleration.
We know from Newton's second law that:
\[ F = m \cdot a \]
where \( a \) is the acceleration. Since acceleration can also be expressed as the derivative of velocity with respect to time, we have:
\[ F = m \cdot \frac{dv}{dt} \]

### Step 2: Substitute the expression for force into the equation.
Given the force \( F(t) = F_0 e^{-bt} \), we can substitute this into the equation:
\[ m \cdot \frac{dv}{dt} = F_0 e^{-bt} \]

### Step 3: Rearrange the equation to isolate \( dv \).
Rearranging gives us:
\[ \frac{dv}{dt} = \frac{F_0}{m} e^{-bt} \]

### Step 4: Integrate both sides.
To find the velocity, we need to integrate both sides with respect to time \( t \):
\[ \int dv = \int \frac{F_0}{m} e^{-bt} dt \]
The left side integrates to \( v \), and the right side can be integrated using the formula for the integral of an exponential function:
\[ v = \frac{F_0}{m} \int e^{-bt} dt = \frac{F_0}{m} \left( -\frac{1}{b} e^{-bt} \right) + C \]
where \( C \) is the constant of integration.

### Step 5: Determine the constant of integration.
At \( t = 0 \), the particle is at rest, so \( v(0) = 0 \):
\[ 0 = -\frac{F_0}{mb} e^{0} + C \implies C = \frac{F_0}{mb} \]

### Step 6: Substitute back to find the expression for velocity.
Now substituting \( C \) back into the equation gives:
\[ v(t) = -\frac{F_0}{mb} e^{-bt} + \frac{F_0}{mb} \]
This can be simplified to:
\[ v(t) = \frac{F_0}{mb} (1 - e^{-bt}) \]

### Step 7: Analyze the behavior of the velocity function.
As \( t \to \infty \), \( e^{-bt} \to 0 \), thus:
\[ v(t) \to \frac{F_0}{mb} \]
This indicates that the velocity approaches a constant value as time increases.

### Conclusion: Identify the correct graph.
The velocity function \( v(t) = \frac{F_0}{mb} (1 - e^{-bt}) \) starts at 0 when \( t = 0 \) and asymptotically approaches \( \frac{F_0}{mb} \) as \( t \) increases. This indicates that the graph of \( v(t) \) is an increasing function that approaches a horizontal asymptote.

### Final Answer:
The correct curve that depicts the speed \( v(t) \) is option B.

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