To solve the problem, we need to analyze the motion of a point under the influence of a retardation proportional to the square of its speed. The retardation is given as \( a v^2 \), where \( a \) is a positive constant and \( v \) is the speed of the point. The initial speed is \( u \), and we want to find the distance covered in time \( t \).
### Step 1: Set up the equation for retardation
The retardation can be expressed as:
\[
\text{Retardation} = -a v^2
\]
This means that the rate of change of speed \( v \) with respect to time \( t \) is given by:
\[
\frac{dv}{dt} = -a v^2
\]
### Step 2: Rearrange the equation
We can rearrange this equation to separate the variables:
\[
\frac{dv}{v^2} = -a \, dt
\]
### Step 3: Integrate both sides
Now, we integrate both sides. The left side will be integrated with respect to \( v \) and the right side with respect to \( t \):
\[
\int \frac{dv}{v^2} = -a \int dt
\]
The integral of \( \frac{1}{v^2} \) is \( -\frac{1}{v} \), and the integral of \( dt \) is \( t \):
\[
-\frac{1}{v} = -at + C
\]
where \( C \) is the constant of integration.
### Step 4: Solve for the constant of integration
To find \( C \), we use the initial condition. At \( t = 0 \), \( v = u \):
\[
-\frac{1}{u} = C
\]
Thus, we can rewrite the equation as:
\[
-\frac{1}{v} = -at - \frac{1}{u}
\]
### Step 5: Simplify the equation
Multiplying through by -1 gives:
\[
\frac{1}{v} = at + \frac{1}{u}
\]
Taking the reciprocal, we find:
\[
v = \frac{1}{\frac{1}{u} + at}
\]
### Step 6: Find the distance covered
To find the distance \( s \) covered in time \( t \), we need to integrate the speed \( v \) with respect to time:
\[
s = \int_0^t v \, dt = \int_0^t \frac{1}{\frac{1}{u} + at} \, dt
\]
### Step 7: Solve the integral
This integral can be solved using a substitution. Let \( x = \frac{1}{u} + at \), then \( dx = a \, dt \) or \( dt = \frac{dx}{a} \). The limits change accordingly:
- When \( t = 0 \), \( x = \frac{1}{u} \)
- When \( t = t \), \( x = \frac{1}{u} + at \)
Thus, the integral becomes:
\[
s = \int_{\frac{1}{u}}^{\frac{1}{u} + at} \frac{1}{x} \frac{dx}{a} = \frac{1}{a} \left[ \ln x \right]_{\frac{1}{u}}^{\frac{1}{u} + at}
\]
Calculating this gives:
\[
s = \frac{1}{a} \left( \ln\left(\frac{1}{u} + at\right) - \ln\left(\frac{1}{u}\right) \right) = \frac{1}{a} \ln\left(\frac{\frac{1}{u} + at}{\frac{1}{u}}\right)
\]
### Final Result
Thus, the distance covered in \( t \) seconds is:
\[
s = \frac{1}{a} \ln\left(1 + u a t\right)
\]
