A body moves in a straight line under the retardation a which is given by `a=bv^2`. If the initial velocity is 2 m/s, the distance covered in time t=2 seconds is

Options

A. b In (4)
B. `1/b In (1+4b)`
C. b In (1-4b)
D. `1/b In (4b-1)`

Answer

To solve the problem of a body moving in a straight line under the retardation given by \( a = -bv^2 \), we will follow these steps:

### Step 1: Understand the given information
We know:
- The acceleration (retardation) is given by \( a = -bv^2 \).
- The initial velocity \( u = 2 \, \text{m/s} \).
- We need to find the distance \( s \) covered in time \( t = 2 \, \text{s} \).

### Step 2: Relate acceleration to velocity
Acceleration can be expressed as:
\[ a = \frac{dv}{dt} = -bv^2 \]
We can rearrange this to separate variables:
\[ \frac{dv}{v^2} = -b \, dt \]

### Step 3: Integrate both sides
Integrate both sides:
\[ \int \frac{dv}{v^2} = \int -b \, dt \]
The left side integrates to:
\[ -\frac{1}{v} = -bt + C \]
where \( C \) is the constant of integration.

### Step 4: Solve for the constant of integration
At \( t = 0 \), the initial velocity \( v = u = 2 \, \text{m/s} \):
\[ -\frac{1}{2} = C \]
So, we have:
\[ -\frac{1}{v} = -bt - \frac{1}{2} \]
or rearranging gives:
\[ \frac{1}{v} = bt + \frac{1}{2} \]

### Step 5: Express velocity in terms of time
Now, we can express \( v \):
\[ v = \frac{1}{bt + \frac{1}{2}} \]

### Step 6: Relate distance to velocity
We know that:
\[ v = \frac{ds}{dt} \]
Thus, we can write:
\[ ds = v \, dt = \frac{1}{bt + \frac{1}{2}} \, dt \]

### Step 7: Integrate to find distance
Now we integrate to find the distance \( s \):
\[ s = \int \frac{1}{bt + \frac{1}{2}} \, dt \]
This integral can be solved using the natural logarithm:
\[ s = \frac{1}{b} \ln \left| bt + \frac{1}{2} \right| + C' \]
where \( C' \) is another constant of integration.

### Step 8: Determine the constant of integration for distance
At \( t = 0 \), \( s = 0 \):
\[ 0 = \frac{1}{b} \ln \left| 0 + \frac{1}{2} \right| + C' \]
Thus, \( C' = -\frac{1}{b} \ln \left( \frac{1}{2} \right) \).

### Step 9: Final expression for distance
Now substituting back, we have:
\[ s = \frac{1}{b} \ln \left| bt + \frac{1}{2} \right| - \frac{1}{b} \ln \left( \frac{1}{2} \right) \]
This simplifies to:
\[ s = \frac{1}{b} \left( \ln \left( bt + \frac{1}{2} \right) - \ln \left( \frac{1}{2} \right) \right) \]
or
\[ s = \frac{1}{b} \ln \left( \frac{bt + \frac{1}{2}}{\frac{1}{2}} \right) \]

### Step 10: Substitute \( t = 2 \)
Now we substitute \( t = 2 \):
\[ s = \frac{1}{b} \ln \left( \frac{2b + \frac{1}{2}}{\frac{1}{2}} \right) = \frac{1}{b} \ln \left( 4b + 1 \right) \]

### Final Result
Thus, the distance covered in 2 seconds is:
\[ s = \frac{1}{b} \ln(4b + 1) \]