An object is subjected to an acceleration `a = 4 + 3v`. It is given that the displacement `S = 0` when `v = 0`. The value of displacement when `v = 3 m//s` is

Options

A. `1 + (4)/(3) ln((13)/(4))`
B. `1 - (4)/(3)ln ((13)/(8))`
C. `1 - (4)/(9) ln ((13)/(4))`
D. `1 + (4)/(9) ln ((13)/(8))`

Explanation

To solve the problem step by step, we will start with the given acceleration and integrate it to find the displacement when the velocity is 3 m/s.

### Step 1: Write down the given acceleration
The acceleration \( a \) is given by:
\[ a = 4 + 3v \]

### Step 2: Relate acceleration to velocity and displacement
We know that acceleration can also be expressed in terms of velocity and displacement:
\[ a = v \frac{dv}{ds} \]
Thus, we can set the two expressions for acceleration equal to each other:
\[ v \frac{dv}{ds} = 4 + 3v \]

### Step 3: Rearrange the equation
Rearranging gives:
\[ v \frac{dv}{ds} = 4 + 3v \implies \frac{dv}{ds} = \frac{4 + 3v}{v} \]
This can be rewritten as:
\[ \frac{dv}{ds} = \frac{4}{v} + 3 \]

### Step 4: Separate variables
We can separate the variables to integrate:
\[ \frac{v}{4 + 3v} dv = ds \]

### Step 5: Integrate both sides
Now we integrate both sides. The left side requires partial fraction decomposition:
\[ \int \frac{v}{4 + 3v} dv = \int ds \]

Using the substitution \( u = 4 + 3v \), we find \( du = 3 dv \) or \( dv = \frac{du}{3} \):
\[ \int \frac{(u - 4)}{u} \cdot \frac{du}{3} = \int ds \]
This simplifies to:
\[ \frac{1}{3} \int (1 - \frac{4}{u}) du = s + C \]
Integrating gives:
\[ \frac{1}{3} \left( u - 4 \ln |u| \right) = s + C \]
Substituting back \( u = 4 + 3v \):
\[ \frac{1}{3} \left( 4 + 3v - 4 \ln |4 + 3v| \right) = s + C \]

### Step 6: Apply initial conditions
We know that when \( v = 0 \), \( s = 0 \):
\[ 0 = \frac{1}{3} (4 - 4 \ln 4) + C \]
Thus:
\[ C = -\frac{4}{3} + \frac{4}{3} \ln 4 \]

### Step 7: Find displacement when \( v = 3 \)
Now we substitute \( v = 3 \) into our equation:
\[ s = \frac{1}{3} \left( 4 + 9 - 4 \ln(4 + 9) \right) + C \]
Calculating:
\[ s = \frac{1}{3} \left( 13 - 4 \ln 13 \right) - \frac{4}{3} + \frac{4}{3} \ln 4 \]
Simplifying gives:
\[ s = \frac{1}{3} \left( 13 - 4 \ln 13 + 4 \ln 4 - 4 \right) \]
\[ s = \frac{1}{3} \left( 9 + 4 \ln \frac{4}{13} \right) \]

### Final Answer
Thus, the value of displacement when \( v = 3 \, \text{m/s} \) is:
\[ s = \frac{1}{3} (9 + 4 \ln \frac{4}{13}) \]