A particle moves in a straight line with acceleration proportional to `x^2`, ( Where x is displacement). The gain of kinetic energy for any displacement is proportional to

Options

A. `x^3`
B. ` x^ (1/3)`
C. `x ^ (2/3)`
D. `x^2`

Answer

To solve the problem, we need to find the relationship between the gain of kinetic energy and the displacement \( x \) when the acceleration is proportional to \( x^2 \).

### Step-by-step Solution:

1. **Understanding the Problem**:
We are given that the acceleration \( a \) is proportional to the square of the displacement \( x \). This can be expressed mathematically as:
\[ a = kx^2 \]
where \( k \) is a constant of proportionality.

2. **Relating Acceleration to Velocity**:
We know that acceleration \( a \) can be expressed as:
\[ a = \frac{dv}{dt} \]
where \( v \) is the velocity. Therefore, we can write:
\[ \frac{dv}{dt} = kx^2 \]

3. **Using Chain Rule**:
To relate \( dv \) and \( dx \), we can use the chain rule:
\[ \frac{dv}{dt} = \frac{dv}{dx} \cdot \frac{dx}{dt} \]
Substituting this into our equation gives:
\[ \frac{dv}{dx} \cdot v = kx^2 \]

4. **Rearranging the Equation**:
We can rearrange this to isolate \( v \):
\[ v \, dv = kx^2 \, dx \]

5. **Integrating Both Sides**:
Now, we integrate both sides:
\[ \int v \, dv = \int kx^2 \, dx \]
The left side integrates to:
\[ \frac{v^2}{2} \]
The right side integrates to:
\[ \frac{kx^3}{3} \]
Thus, we have:
\[ \frac{v^2}{2} = \frac{kx^3}{3} + C \]
where \( C \) is the constant of integration.

6. **Finding Kinetic Energy**:
The kinetic energy \( KE \) of the particle is given by:
\[ KE = \frac{1}{2} mv^2 \]
From our previous result, we can express \( v^2 \) in terms of \( x \):
\[ v^2 = \frac{2}{3} kx^3 + 2C \]
Therefore, the kinetic energy becomes:
\[ KE \propto v^2 \propto x^3 \]

7. **Conclusion**:
The gain in kinetic energy as the displacement changes is proportional to \( x^3 \).

### Final Answer:
The gain of kinetic energy for any displacement is proportional to \( x^3 \).