The motion of a body falling from rest in a resisting medium is described by the equation `(dv)//(dt)=a-bv` where a and b are constant . The velocity at any time `t` is given by

Options

A. `a(1-b^(2t))`
B. `(a)/(b)(1-e^(-bt))`
C. `abe^(-1)`
D. `ab^(2)(1-t)`

Answer

To find the velocity of a body falling from rest in a resisting medium described by the equation \(\frac{dv}{dt} = a - bv\), we can follow these steps:

### Step 1: Rearrange the equation
We start with the equation:
\[ \frac{dv}{dt} = a - bv \]
We can rearrange this to isolate \(dv\) on one side:
\[ \frac{dv}{a - bv} = dt \]

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

### Step 3: Solve the left integral
The left integral can be solved using the natural logarithm:
\[ \int \frac{dv}{a - bv} = -\frac{1}{b} \ln |a - bv| + C_1 \]
The right side integrates to:
\[ \int dt = t + C_2 \]

### Step 4: Combine the results
Setting the two integrals equal gives:
\[ -\frac{1}{b} \ln |a - bv| = t + C \]
where \(C = C_2 - C_1\).

### Step 5: Solve for \(v\)
To solve for \(v\), we first multiply through by \(-b\):
\[ \ln |a - bv| = -bt - bC \]
Exponentiating both sides results in:
\[ |a - bv| = e^{-bt - bC} \]

### Step 6: Remove the absolute value
Since the body is falling and \(v\) will be less than \(a/b\) (as it approaches terminal velocity), we can drop the absolute value:
\[ a - bv = e^{-bt - bC} \]

### Step 7: Solve for \(v\)
Rearranging gives:
\[ bv = a - e^{-bt - bC} \]
Thus,
\[ v = \frac{a}{b} - \frac{e^{-bt - bC}}{b} \]

### Step 8: Determine the constant
Since the body starts from rest, when \(t = 0\), \(v = 0\):
\[ 0 = \frac{a}{b} - \frac{e^{-bC}}{b} \]
This implies:
\[ e^{-bC} = a \]
Substituting this back into our equation for \(v\):
\[ v = \frac{a}{b} - \frac{a e^{-bt}}{b} \]
This simplifies to:
\[ v = \frac{a}{b}(1 - e^{-bt}) \]

### Final Result
The velocity at any time \(t\) is given by:
\[ v(t) = \frac{a}{b}(1 - e^{-bt}) \]