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 \( a = 4 - 2v \), we can follow these steps:

### Step 1: Write the relationship between acceleration and velocity
We know that acceleration \( a \) can also be expressed as the derivative of velocity with respect to time:
\[ a = \frac{dv}{dt} \]
Thus, we can set up the equation:
\[ \frac{dv}{dt} = 4 - 2v \]

### Step 2: Rearrange the equation
To solve this differential equation, we can rearrange it:
\[ \frac{dv}{4 - 2v} = dt \]

### Step 3: Integrate both sides
Now, we will integrate both sides. The left side requires a substitution:
\[ \int \frac{dv}{4 - 2v} = \int dt \]
The left side can be simplified by factoring out the constant:
\[ \int \frac{1}{2} \cdot \frac{1}{2 - v} dv = \int dt \]
This gives us:
\[ \frac{1}{2} \ln |4 - 2v| = t + C \]
where \( C \) is the constant of integration.

### Step 4: Solve for \( v \)
To solve for \( v \), we can exponentiate both sides:
\[ |4 - 2v| = e^{2(t + C)} = e^{2t} \cdot e^{2C} \]
Let \( K = e^{2C} \), then:
\[ 4 - 2v = K e^{2t} \]
Solving for \( v \):
\[ 2v = 4 - K e^{2t} \]
\[ v = 2 - \frac{K}{2} e^{2t} \]

### Step 5: Determine the maximum velocity
To find the maximum velocity, we need to consider the limit as \( t \) approaches infinity. As \( t \to \infty \), the term \( e^{2t} \) grows very large, and thus \( \frac{K}{2} e^{2t} \) will dominate unless \( K \) is zero. For maximum velocity, we want the negative term to vanish:
\[ v_{\text{max}} = 2 - 0 = 2 \]

### Conclusion
Thus, the maximum velocity \( v_{\text{max}} \) is:
\[ \boxed{2 \text{ m/s}} \]