A body is in straight line motion with an acceleration given by `a=32-4v`. The initial conditions are at `t=0, v=4`. Find the velocity when `t=ln 2`:

Options

A. `15//2`
B. `17//2`
C. `23//4`
D. `31//4`

Answer

To solve the problem, we need to find the velocity of a body in straight line motion given its acceleration as a function of velocity. The acceleration is defined as:

\[ a = 32 - 4v \]

where \( v \) is the velocity. We also know the initial conditions: at \( t = 0 \), \( v = 4 \). We want to find the velocity when \( t = \ln(2) \).

### Step 1: Set up the relationship between acceleration and velocity
We know that acceleration \( a \) can be expressed as the derivative of velocity with respect to time:

\[ a = \frac{dv}{dt} \]

Thus, we can write:

\[ \frac{dv}{dt} = 32 - 4v \]

### Step 2: Rearrange the equation for integration
We can rearrange this equation to separate the variables \( v \) and \( t \):

\[ \frac{dv}{32 - 4v} = dt \]

### Step 3: Integrate both sides
Now we will integrate both sides. The left side will be integrated with respect to \( v \) and the right side with respect to \( t \):

\[ \int \frac{dv}{32 - 4v} = \int dt \]

The left side can be simplified by factoring out the constant:

\[ \int \frac{1}{4} \cdot \frac{dv}{8 - v} = \int dt \]

### Step 4: Solve the integral
The integral on the left side can be solved as follows:

\[ \frac{1}{4} \ln |32 - 4v| = t + C \]

where \( C \) is the constant of integration.

### Step 5: Apply initial conditions
We need to find the constant \( C \) using the initial conditions \( t = 0 \) and \( v = 4 \):

\[ \frac{1}{4} \ln |32 - 4(4)| = 0 + C \]

This simplifies to:

\[ \frac{1}{4} \ln |32 - 16| = C \]
\[ C = \frac{1}{4} \ln 16 = \frac{1}{4} \cdot 4 \ln 2 = \ln 2 \]

### Step 6: Substitute back to find velocity
Now we substitute \( C \) back into the equation:

\[ \frac{1}{4} \ln |32 - 4v| = t + \ln 2 \]

### Step 7: Solve for \( v \) when \( t = \ln 2 \)
Now we set \( t = \ln 2 \):

\[ \frac{1}{4} \ln |32 - 4v| = \ln 2 + \ln 2 \]
\[ \frac{1}{4} \ln |32 - 4v| = 2 \ln 2 \]
\[ \ln |32 - 4v| = 8 \ln 2 \]
\[ |32 - 4v| = 2^8 = 256 \]

### Step 8: Solve for \( v \)
Now we solve for \( v \):

1. \( 32 - 4v = 256 \)
\[ -4v = 256 - 32 \]
\[ -4v = 224 \]
\[ v = -56 \] (not physically meaningful)

2. \( 32 - 4v = -256 \)
\[ -4v = -256 - 32 \]
\[ -4v = -288 \]
\[ v = 72 \]

Thus, the velocity when \( t = \ln 2 \) is:

\[ v = 72 \]

### Final Answer:
The velocity when \( t = \ln 2 \) is \( v = 72 \).

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