The velocity v of a particle moving along x - axis varies with ist position (x) as `v=alpha sqrtx`, where `alpha` is a constant. Which of the following graph represents the variation of its acceleration (a) with time (t)?

Options

A.
B.
C.
D.

Answer

To solve the problem, we need to find the relationship between acceleration \( a \) and time \( t \) for a particle whose velocity \( v \) varies with position \( x \) as \( v = \alpha \sqrt{x} \).

### Step-by-Step Solution:

1. **Understanding the Given Relation**:
The velocity of the particle is given by:
\[ v = \alpha \sqrt{x} \]
where \( \alpha \) is a constant.

2. **Finding Acceleration**:
Acceleration \( a \) is defined as the rate of change of velocity with respect to time:
\[ a = \frac{dv}{dt} \]
We can use the chain rule to express this in terms of position:
\[ a = \frac{dv}{dx} \cdot \frac{dx}{dt} \]
Here, \( \frac{dx}{dt} \) is the velocity \( v \).

3. **Calculating \( \frac{dv}{dx} \)**:
We need to differentiate \( v \) with respect to \( x \):
\[ v = \alpha \sqrt{x} \]
Differentiating:
\[ \frac{dv}{dx} = \frac{\alpha}{2\sqrt{x}} \]

4. **Substituting Back into the Acceleration Formula**:
Now substituting \( \frac{dv}{dx} \) and \( v \) into the acceleration formula:
\[ a = v \cdot \frac{dv}{dx} = \alpha \sqrt{x} \cdot \frac{\alpha}{2\sqrt{x}} = \frac{\alpha^2}{2} \]
This shows that the acceleration \( a \) is a constant value, independent of \( x \).

5. **Relating Acceleration to Time**:
Since the acceleration \( a \) is constant, it does not change with time. Therefore, the graph representing the variation of acceleration \( a \) with time \( t \) will be a horizontal line.

6. **Conclusion**:
The correct graph representing the variation of acceleration \( a \) with time \( t \) is a horizontal line, indicating that acceleration is constant.

### Final Answer:
The graph that represents the variation of acceleration \( a \) with time \( t \) is a horizontal line (constant acceleration).

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