The motion of a particle moving along x-axis is represented by the equation `(dv)/(dt)=6-3v`, where v is in m/s and t is in second. If the particle is at rest at t = 0 , then

Options

A. The speed of the particle in 2 m/s when the acceleration of particle is zero
B. After a long time the particle moves with a constant velocity of 2 m/s
C. The magnitude of final acceleration is `6m//s^(2)`
D. The speed is 0.1 m/s, when the acceleration is half of its initial value

Answer

To solve the problem, we start with the given differential equation that describes the motion of the particle:

\[ \frac{dv}{dt} = 6 - 3v \]

### Step 1: Identify the initial condition
The problem states that the particle is at rest at \( t = 0 \). This means:
\[ v(0) = 0 \, \text{m/s} \]

### Step 2: Rearrange the equation
We can rearrange the differential equation to separate variables:
\[ \frac{dv}{6 - 3v} = dt \]

### Step 3: Integrate both sides
Now, we integrate both sides. The left side requires a logarithmic integration:
\[ \int \frac{dv}{6 - 3v} = \int dt \]

The left side can be integrated using the substitution \( u = 6 - 3v \), which gives \( du = -3 dv \) or \( dv = -\frac{du}{3} \):
\[ -\frac{1}{3} \int \frac{1}{u} du = \int dt \]
This results in:
\[ -\frac{1}{3} \ln |6 - 3v| = t + C \]

### Step 4: Solve for \( v \)
To solve for \( v \), we first multiply through by -3:
\[ \ln |6 - 3v| = -3t - 3C \]
Exponentiating both sides gives:
\[ 6 - 3v = e^{-3t - 3C} \]
Let \( K = e^{-3C} \), then:
\[ 6 - 3v = Ke^{-3t} \]
Rearranging gives:
\[ 3v = 6 - Ke^{-3t} \]
Thus:
\[ v = 2 - \frac{K}{3} e^{-3t} \]

### Step 5: Apply the initial condition
Using the initial condition \( v(0) = 0 \):
\[ 0 = 2 - \frac{K}{3} e^{0} \]
This simplifies to:
\[ 0 = 2 - \frac{K}{3} \implies K = 6 \]

### Step 6: Substitute back to find \( v(t) \)
Substituting \( K \) back into the equation for \( v \):
\[ v = 2 - 2e^{-3t} \]

### Step 7: Analyze the behavior as \( t \to \infty \)
As \( t \) approaches infinity, \( e^{-3t} \) approaches 0:
\[ v(t) \to 2 \, \text{m/s} \]

### Conclusion
The speed of the particle approaches \( 2 \, \text{m/s} \) as \( t \) becomes very large, confirming that the particle moves with a constant velocity of \( 2 \, \text{m/s} \) in the long term.