A particle is projected from origin `x`-axis with velocity `V_0` such that it suffers retardation of magnitude given by `Kx^3` (where k is a positive constant).The stopping distance of particle is

Options

A. `((3V_0^2)/(4k))^(1/3)`
B. `((2V_0^2)/(k))^(1/4)`
C. `(V_0^2/(2k))^(1/4)`
D. `((2V_0^2)/(3k))^(1/3)`

Answer

To solve the problem of finding the stopping distance of a particle projected from the origin with an initial velocity \( V_0 \) and experiencing a retardation of magnitude \( Kx^3 \), we can follow these steps:

### Step 1: Understand the relationship between acceleration and retardation
The retardation \( a \) is given by:
\[ a = -Kx^3 \]
This means that the acceleration is negative and depends on the position \( x \).

### Step 2: Use the relationship between acceleration, velocity, and position
We can use the relationship:
\[ a = v \frac{dv}{dx} \]
Substituting for \( a \):
\[ v \frac{dv}{dx} = -Kx^3 \]

### Step 3: Rearrange the equation for integration
Rearranging gives:
\[ v \, dv = -Kx^3 \, dx \]

### Step 4: Integrate both sides
Now we integrate both sides. For the left side, we integrate with respect to \( v \) from \( V_0 \) to \( 0 \) (the final velocity when the particle stops), and for the right side, we integrate with respect to \( x \) from \( 0 \) to \( s \) (the stopping distance):
\[ \int_{V_0}^{0} v \, dv = -K \int_{0}^{s} x^3 \, dx \]

### Step 5: Perform the integrations
The left side becomes:
\[ \frac{v^2}{2} \bigg|_{V_0}^{0} = 0 - \frac{V_0^2}{2} = -\frac{V_0^2}{2} \]
The right side becomes:
\[ -K \left( \frac{x^4}{4} \bigg|_{0}^{s} \right) = -K \left( \frac{s^4}{4} - 0 \right) = -\frac{Ks^4}{4} \]

### Step 6: Set the equations equal to each other
Setting the results of the integrations equal gives:
\[ -\frac{V_0^2}{2} = -\frac{Ks^4}{4} \]

### Step 7: Solve for \( s \)
Removing the negative signs and rearranging gives:
\[ \frac{V_0^2}{2} = \frac{Ks^4}{4} \]
Multiplying both sides by 4:
\[ 2V_0^2 = Ks^4 \]
Now, solving for \( s^4 \):
\[ s^4 = \frac{2V_0^2}{K} \]
Taking the fourth root gives:
\[ s = \left( \frac{2V_0^2}{K} \right)^{1/4} \]

### Final Answer
Thus, the stopping distance \( s \) of the particle is:
\[ s = \left( \frac{2V_0^2}{K} \right)^{1/4} \]

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