A point moves in a straight line under the retardation `av^(2)`, where `'a'` is a positive constant and `v` is speed. If the initial speed is `u`, the distance covered in `'t'` seconds is `:`

Options

A. `aut`
B. `(1)/(a) ln (aut)`
C. `(1)/(a)ln(1+aut)`
D. `a ln (aut)`

Explanation

To solve the problem, we need to determine the distance covered by a point moving in a straight line under the retardation \( a v^2 \), where \( a \) is a positive constant and \( v \) is the speed. The initial speed is \( u \), and we want to find the distance covered in \( t \) seconds.

### Step-by-step Solution:

1. **Understanding Retardation**:
The retardation can be expressed as:
\[ a = -a v^2 \]
This implies that the acceleration \( \frac{dv}{dt} \) is negative, indicating a decrease in speed.

2. **Setting Up the Differential Equation**:
We can rewrite the equation as:
\[ \frac{dv}{dt} = -a v^2 \]
Rearranging gives:
\[ \frac{dv}{v^2} = -a \, dt \]

3. **Integrating Both Sides**:
We will integrate both sides. The left side integrates from \( u \) (initial speed) to \( v \) (speed at time \( t \)), and the right side integrates from \( 0 \) to \( t \):
\[ \int_{u}^{v} \frac{dv}{v^2} = -a \int_{0}^{t} dt \]
The integral of \( \frac{1}{v^2} \) is \( -\frac{1}{v} \):
\[ -\frac{1}{v} + \frac{1}{u} = -at \]

4. **Solving for \( v \)**:
Rearranging the equation gives:
\[ \frac{1}{u} - \frac{1}{v} = at \]
This leads to:
\[ \frac{1}{v} = \frac{1}{u} - at \]
Therefore:
\[ v = \frac{u}{1 + a u t} \]

5. **Finding the Distance**:
The distance \( x \) covered in time \( t \) can be expressed as:
\[ v = \frac{dx}{dt} \]
Thus:
\[ dx = v \, dt = \frac{u}{1 + a u t} \, dt \]
Now, we integrate both sides:
\[ x = \int_{0}^{t} \frac{u}{1 + a u t} \, dt \]

6. **Integrating**:
The integral can be computed as follows:
\[ x = u \int_{0}^{t} \frac{1}{1 + a u t} \, dt \]
The integral \( \int \frac{1}{1 + a u t} \, dt \) is:
\[ \frac{1}{a u} \ln(1 + a u t) \]
Therefore:
\[ x = \frac{u}{a u} \left[ \ln(1 + a u t) \right]_{0}^{t} \]
Evaluating the limits gives:
\[ x = \frac{1}{a} \ln(1 + a u t) \]

### Final Result:
The distance covered in \( t \) seconds is:
\[ x = \frac{1}{a} \ln(1 + a u t) \]