A body of mass m = 4 kg starts moving with velocity `v_(0)` in a straight line is such a way that on the body work is being done at the rate which is proportional to the square of velocity as given by `P = beta v^(2)` where `beta = ( 0.693)/( 2)`. Find the time elapsed in seconds before velocity of body is doubled.

Answer

To solve the problem, we need to find the time elapsed before the velocity of the body is doubled. We are given that the power \( P \) is proportional to the square of the velocity, expressed as:

\[ P = \beta v^2 \]

where \( \beta = \frac{0.693}{2} \).

### Step-by-Step Solution:

1. **Understanding Power and Work Done**:
The power \( P \) can be expressed in terms of force and velocity:
\[ P = F \cdot v \]
where \( F \) is the force acting on the body. The force can also be expressed in terms of mass \( m \) and acceleration \( a \):
\[ F = m \frac{dv}{dt} \]
Therefore, we can write:
\[ P = m \frac{dv}{dt} \cdot v \]

2. **Setting the Equations Equal**:
From the power equation, we have:
\[ \beta v^2 = m \frac{dv}{dt} \cdot v \]
Dividing both sides by \( v \) (assuming \( v \neq 0 \)):
\[ \beta v = m \frac{dv}{dt} \]

3. **Rearranging the Equation**:
Rearranging gives us:
\[ \frac{dv}{dt} = \frac{\beta}{m} v \]

4. **Separating Variables**:
We can separate the variables to integrate:
\[ \frac{dv}{v} = \frac{\beta}{m} dt \]

5. **Integrating Both Sides**:
Integrating both sides, we consider the initial velocity \( v_0 \) and the final velocity \( 2v_0 \):
\[ \int_{v_0}^{2v_0} \frac{1}{v} dv = \int_{0}^{t} \frac{\beta}{m} dt \]
This results in:
\[ \ln(2v_0) - \ln(v_0) = \frac{\beta}{m} t \]
Simplifying gives:
\[ \ln(2) = \frac{\beta}{m} t \]

6. **Substituting Values**:
Now substituting \( \beta = \frac{0.693}{2} \) and \( m = 4 \):
\[ \ln(2) = \frac{0.693/2}{4} t \]
Simplifying further:
\[ \ln(2) = \frac{0.693}{8} t \]

7. **Solving for Time \( t \)**:
Rearranging to solve for \( t \):
\[ t = \frac{8 \ln(2)}{0.693} \]
Since \( \ln(2) \approx 0.693 \):
\[ t = 8 \text{ seconds} \]

### Final Answer:
The time elapsed before the velocity of the body is doubled is **8 seconds**.