The motion of a body falling from rest in a medium is given by equation,`(dv)/(dt)=1-2v`. The velocity at any time t is

Options

A. `(1-e^t)`
B. `(1-e^(-t))`
C. `1/2(1-e^t)`
D. `1/2(1-e^(-2t))`

Answer

To solve the problem, we will follow these steps:

### Step 1: Write down the given equation
The equation provided is:
\[ \frac{dv}{dt} = 1 - 2v \]

### Step 2: Separate the variables
We want to separate the variables \(v\) and \(t\) so that we can integrate. We can rearrange the equation as follows:
\[ \frac{dv}{1 - 2v} = dt \]

### Step 3: Integrate both sides
Next, we will integrate both sides. The left side will be integrated with respect to \(v\) and the right side with respect to \(t\):
\[ \int \frac{dv}{1 - 2v} = \int dt \]

### Step 4: Solve the integrals
The integral on the left side can be solved using the natural logarithm:
\[ -\frac{1}{2} \ln|1 - 2v| = t + C \]
where \(C\) is the constant of integration.

### Step 5: Solve for \(v\)
To isolate \(v\), we first multiply both sides by -2:
\[ \ln|1 - 2v| = -2(t + C) \]
Now, we can exponentiate both sides:
\[ 1 - 2v = e^{-2(t + C)} \]
This can be rewritten as:
\[ 1 - 2v = e^{-2t} e^{-2C} \]
Let \(k = e^{-2C}\) (a constant), we have:
\[ 1 - 2v = k e^{-2t} \]

### Step 6: Isolate \(v\)
Now, we can solve for \(v\):
\[ 2v = 1 - k e^{-2t} \]
\[ v = \frac{1 - k e^{-2t}}{2} \]

### Step 7: Determine the constant \(k\)
Since the body is falling from rest, we know that at \(t = 0\), \(v = 0\):
\[ 0 = \frac{1 - k}{2} \]
This implies:
\[ 1 - k = 0 \quad \Rightarrow \quad k = 1 \]

### Step 8: Substitute \(k\) back into the equation
Now substituting \(k\) back into the equation for \(v\):
\[ v = \frac{1 - e^{-2t}}{2} \]

### Final Answer
Thus, the velocity at any time \(t\) is:
\[ v(t) = \frac{1 - e^{-2t}}{2} \]

---