The intial velocity of a particle moving along x axis is u (at t = 0 and x = 0) and its acceleration a is given by a = kx. Which of the following equation is correct between its velocity (v) and position (x)?

Options

A. `v^(2)=y^(2)=2kx`
B. `v^(2)=u^(2)+2kx^(2)`
C. `v^(2)=u^(2)+kx^(2)`
D. `v^(2)+u^(2)=2kx`

Answer

To solve the problem, we need to derive the relationship between the velocity \( v \) and the position \( x \) of a particle moving along the x-axis, given that its acceleration \( a \) is proportional to its position \( x \) (i.e., \( a = kx \)).

### Step-by-step Solution:

1. **Understand the relationship between acceleration, velocity, and position**:
We know that acceleration \( a \) can be expressed as:
\[ a = \frac{dv}{dt} \]
Additionally, we can express \( a \) in terms of velocity and position using the chain rule:
\[ a = v \frac{dv}{dx} \]

2. **Set up the equation**:
Given that \( a = kx \), we can equate the two expressions for acceleration:
\[ v \frac{dv}{dx} = kx \]

3. **Rearrange the equation**:
We can rearrange this equation to separate variables:
\[ v \, dv = kx \, dx \]

4. **Integrate both sides**:
Now, we integrate both sides. The left side integrates with respect to \( v \) and the right side with respect to \( x \):
\[ \int v \, dv = \int kx \, dx \]
This gives us:
\[ \frac{v^2}{2} = \frac{kx^2}{2} + C \]
where \( C \) is the constant of integration.

5. **Apply the initial conditions**:
We know from the problem that at \( t = 0 \) and \( x = 0 \), the initial velocity \( v = u \). We can use this to find \( C \):
\[ \frac{u^2}{2} = \frac{k(0)^2}{2} + C \]
This simplifies to:
\[ C = \frac{u^2}{2} \]

6. **Substitute back to find the relationship**:
Now substituting \( C \) back into our integrated equation, we have:
\[ \frac{v^2}{2} = \frac{kx^2}{2} + \frac{u^2}{2} \]
Multiplying through by 2 to eliminate the fractions gives us:
\[ v^2 = u^2 + kx^2 \]

### Final Result:
The correct equation relating velocity \( v \) and position \( x \) is:
\[ v^2 = u^2 + kx^2 \]