A particle moving in the positive x-direction has initial velocity `v_(0)`. The particle undergoes retardation `kv^(2)` , where vis its instantaneous velocity. The velocity of the particle as a function of time is given by:

Options

A. `v= v_(0) //( 1+ kv_(0) t)`
B. `v= ( 2v_(0))/(1+kt)`
C. `v = ( v_(0))/(kt)`
D. `v = (v_(0))/((1+k^(2) v_(0^(2) t)))`

Answer

To find the velocity of a particle moving in the positive x-direction with an initial velocity \( v_0 \) and undergoing a retardation of \( kv^2 \), we can follow these steps:

### Step 1: Set up the equation for retardation
The retardation is given by \( -kv^2 \). According to Newton's second law, we can express this as:
\[ \frac{dv}{dt} = -kv^2 \]

### Step 2: Rearrange the equation
We can rearrange the equation to separate the variables:
\[ \frac{dv}{v^2} = -k \, dt \]

### Step 3: Integrate both sides
Now, we will integrate both sides. The left side integrates to:
\[ \int \frac{dv}{v^2} = -\frac{1}{v} \]
And the right side integrates to:
\[ \int -k \, dt = -kt + C \]
where \( C \) is the constant of integration.

### Step 4: Combine the results of the integration
Combining the results from the integration gives us:
\[ -\frac{1}{v} = -kt + C \]

### Step 5: Solve for the constant of integration
To find the constant \( C \), we use the initial condition. At \( t = 0 \), \( v = v_0 \):
\[ -\frac{1}{v_0} = C \]
Thus, we can substitute \( C \) back into our equation:
\[ -\frac{1}{v} = -kt - \frac{1}{v_0} \]

### Step 6: Rearrange to find \( v \)
Rearranging the equation gives:
\[ \frac{1}{v} = kt + \frac{1}{v_0} \]
Taking the reciprocal of both sides, we find:
\[ v = \frac{v_0}{1 + kv_0 t} \]

### Final Result
Thus, the velocity of the particle as a function of time is:
\[ v(t) = \frac{v_0}{1 + kv_0 t} \]

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