The acceleration of a particle moving along a straight line at any time t is given by a = 4 - 2v, where v is the speed of particle at any time t The maximum velocity is

Options

A. 4 m/s
B. 2 m/s
C. 6 m/s
D. Infinity

Answer

To find the maximum velocity of a particle moving along a straight line with the given acceleration \( a = 4 - 2v \), where \( v \) is the speed of the particle, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the relationship between acceleration and velocity**:
The acceleration \( a \) can be expressed in terms of velocity \( v \) as:
\[ a = \frac{dv}{dt} = 4 - 2v \]

2. **Rearrange the equation**:
Rearranging the equation gives:
\[ \frac{dv}{4 - 2v} = dt \]

3. **Integrate both sides**:
We will integrate the left side with respect to \( v \) and the right side with respect to \( t \):
\[ \int \frac{dv}{4 - 2v} = \int dt \]
The left side can be simplified using substitution. Let \( u = 4 - 2v \), then \( du = -2dv \) or \( dv = -\frac{du}{2} \):
\[ \int \frac{-\frac{1}{2} du}{u} = \int dt \]
This gives:
\[ -\frac{1}{2} \ln |4 - 2v| = t + C \]

4. **Solve for \( v \)**:
Rearranging the equation, we get:
\[ \ln |4 - 2v| = -2t - 2C \]
Exponentiating both sides results in:
\[ 4 - 2v = e^{-2t - 2C} \]
Let \( K = e^{-2C} \), then:
\[ 4 - 2v = Ke^{-2t} \]
Thus:
\[ 2v = 4 - Ke^{-2t} \]
Therefore:
\[ v = 2 - \frac{K}{2} e^{-2t} \]

5. **Find the maximum velocity**:
As \( t \) approaches infinity, \( e^{-2t} \) approaches 0. Thus:
\[ v_{\text{max}} = 2 - \frac{K}{2} \cdot 0 = 2 \]

6. **Conclusion**:
The maximum velocity of the particle is:
\[ v_{\text{max}} = 2 \, \text{m/s} \]