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 step by step, we will analyze the motion of a particle subjected to a time-dependent force.

### Step 1: Identify the Force
The force acting on the particle is given by:
\[ F(t) = F_0 e^{-bt} \]
where \( F_0 \) is a constant and \( b \) is a decay constant.

### Step 2: Relate Force to Acceleration
According to Newton's second law, the force is related to mass and acceleration:
\[ F = m \cdot a \]
where \( a = \frac{dv}{dt} \) (acceleration is the rate of change of velocity). Thus, we can write:
\[ m \cdot \frac{dv}{dt} = F_0 e^{-bt} \]

### Step 3: Rearranging the Equation
Rearranging the equation gives:
\[ \frac{dv}{dt} = \frac{F_0}{m} e^{-bt} \]

### Step 4: Integrate to Find Velocity
To find the velocity \( v(t) \), 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 \( v(t) \), and the right side can be integrated as follows:
\[ v(t) = \frac{F_0}{m} \int e^{-bt} dt \]

The integral of \( e^{-bt} \) is:
\[ \int e^{-bt} dt = -\frac{1}{b} e^{-bt} + C \]
where \( C \) is the constant of integration.

### Step 5: Apply Initial Conditions
At \( t = 0 \), the particle is at rest, so \( v(0) = 0 \):
\[ v(0) = \frac{F_0}{m} \left(-\frac{1}{b} e^{-b \cdot 0} + C\right) = 0 \]
This simplifies to:
\[ C = \frac{F_0}{mb} \]

### Step 6: Final Expression for Velocity
Substituting \( C \) back into the equation for \( v(t) \):
\[ v(t) = \frac{F_0}{m} \left(-\frac{1}{b} e^{-bt} + \frac{F_0}{mb}\right) \]
This simplifies to:
\[ v(t) = \frac{F_0}{mb} (1 - e^{-bt}) \]

### Step 7: Analyze the Velocity Function
The velocity function \( v(t) = \frac{F_0}{mb} (1 - e^{-bt}) \) indicates that as \( t \) approaches infinity, \( e^{-bt} \) approaches zero, and thus:
\[ v(t) \to \frac{F_0}{mb} \]
This means the velocity approaches a constant value as time increases.

### Step 8: Sketch the Velocity Curve
The velocity starts from zero at \( t = 0 \) and asymptotically approaches \( \frac{F_0}{mb} \) as \( t \) increases. The curve will be an increasing function that levels off, resembling an exponential growth curve.

### Final Answer
The speed \( v(t) \) is depicted by a curve that starts at the origin (0,0) and approaches the horizontal asymptote \( \frac{F_0}{mb} \) as \( t \) increases.

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